Harmonics solutions

Read the wave number from the phase: k=8 rad/m. The traveling-wave amplitude is irrelevant to the node spacing.

Why: The spatial phase coefficient determines wavelength and node spacing; amplitude does not change where the nodes sit.

Adjacent standing-wave nodes are separated by λ/2=π/k=π/8 m.

Why: A standing wave has successive nodes half a wavelength apart, giving π/k between neighbors.

Four nodes including the two fixed ends enclose three node-to-node intervals, so this is harmonic n=3.

Why: Count intervals rather than nodes: four total nodes make three half-wave segments between the fixed ends.

Therefore L=3π/8 m=1.17810 m. The nodes are at x=0, π/8, 2π/8 and 3π/8 m.

Why: Add those three equal node spacings to obtain the string length and the corresponding node positions.

For an ideal spring T(m)=2π√(m/k). Use the same k for both attached masses.

Why: Changing the attached mass leaves the same spring stiffness, so both ideal periods share the same k.

The measured difference is 8 s=2π(√(10 kg)−√(5 kg))/√k.

Why: Subtract the two mass-dependent periods and equate that difference to the measured eight seconds.

Solving gives k=[2π(√10−√5)/8]² N/m=0.529174 N/m.

Why: Isolate √k in the difference equation and square it to find the spring constant.

The periods are T₅=19.3137 s and T₁₀=27.3137 s, whose difference is 8.0000 s. The very small stiffness explains the long periods.

Why: Recalculate both periods to verify their difference; a very soft spring is consistent with these long oscillations.

A fixed-end normal mode satisfies fₙ=nv/(2L), with integer n≥1. Its node count is n+1 including the ends.

Why: The fixed ends force an integer number of half-wavelengths along the string, with one more node than that integer.

Convert L=85 cm=0.85 m and calculate n=2Lf/v=2(0.85)(82.41)/140=1.00069286.

Why: Convert length to meters and solve the normal-mode relation for its candidate harmonic number.

An exact ideal normal mode cannot have a fractional n. The fundamental predicted from the rounded L and v is 140/(2×0.85)=82.3529 Hz.

Why: The near-one result suggests the fundamental, but a fractional harmonic cannot satisfy an exact fixed-end mode.

For the intended fundamental n=1, the exact speed would be v=2Lf=140.097 m/s, which rounds to 140 m/s.

Why: Calculate the speed required by an exact fundamental and check it rounds to the stated approximate speed.

Thus the rounded-data interpretation is the fundamental, with two endpoint nodes and no internal node. If all source numbers are treated as exact, they are inconsistent.

Why: Rounded measurements support two endpoint nodes; treating every given as exact would reject a single common normal mode.

For vertical displacement from static equilibrium, gravity is absorbed into the equilibrium position. Both springs change length by the same displacement, so k_eff=k₁+k₂.

Why: Use displacement from the hanging equilibrium: gravity sets that position while the two spring restoring forces add.

The period gives k_eff=m(2π/T)²=10(2π/4)²=24.6740 N/m.

Why: The claimed period fixes the required total stiffness through the ideal spring-mass period relation.

Subtract the known stiffness: k₂=24.6740−40=−15.3260 N/m.

Why: Subtract the known spring stiffness to isolate the formal stiffness of the second spring.

An ordinary stable spring has positive stiffness, so no physical k₂ of the stated type exists. Even with the second spring absent, the longest possible period is 2π√(10/40)=π s≈3.142 s; adding a positive spring only shortens it.

Why: The required negative stiffness is outside the ordinary-spring model; any positive added stiffness would shorten an already shorter period.

From the given equation, k=4 rad/m and ω=8 rad/s. The component traveling-wave speed is v=ω/k=2 m/s; the standing pattern itself has stationary nodes.

Why: Read spatial and temporal phase rates from the equation; their ratio is traveling-wave speed, not the speed of its stationary nodes.

The equation gives λ=2π/k=π/2 m and T=2π/ω=π/4 s≈0.785398 s, not 5 s.

Why: Turn k and ω into wavelength and period to test the independently stated five-second period.

Four total nodes on a fixed-end string require n=3 and L=3π/k=3π/4 m≈2.35619 m. For the stated L=1.118 m, sin(4L)=sin(4.472) is not zero, so that endpoint is not fixed under this equation.

Why: Four total fixed-end nodes demand three half-wavelengths, and a fixed right endpoint must have zero spatial factor.

Alternatively, retain L=1.118 m, four fixed-end nodes and T=5 s. Then λ=2L/3=0.745333 m and v=λ/T=0.149067 m/s.

Why: If only the length, node count and period are kept, they define a different wavelength and component speed.

The equation-based result 2 m/s and the geometry-and-period result 0.149067 m/s describe different models. All of the source givens cannot hold simultaneously.

Why: The two internally defined speeds differ, so the full set of source conditions cannot describe one standing wave.

The gravitational torque is τ=−mgh sinθ. For small θ in radians, Iθ̈≈−mghθ and T=2π√[I/(mgh)].

Why: Gravity’s torque depends on the center-of-mass lever arm; the small-angle sine approximation makes the rotation harmonic.

The centre-of-mass distance is h=L/3=2/3 m. This is not the same as the full stated length.

Why: The restoring torque uses the pivot-to-center distance, which is only one-third of the stated full length.

Rearrange the period relation: m=4π²I/(ghT²).

Why: Isolate m in the physical-pendulum period equation while retaining the supplied pivot inertia.

Insert the data: m=4π²(20)/[9.8(2/3)(5²)]=4.83409 kg.

Why: Insert h, I and the measured period with g to obtain the mass implied by that formula.

Substitution into 2π√[20/(4.83409×9.8×2/3)] recovers 5.000 s. The dimensions are kg·m²/[(m/s²)×m×s²]=kg.

Why: Check both the recovered period and mass units; this verifies the algebra while retaining the result’s conditional geometry assumption.

At the bottom, the gravitational torque about the pivot is zero, so angular acceleration and tangential acceleration are zero. The centre and tip still have radial centripetal acceleration toward the pivot.

Why: At the bottom gravity passes through the pivot, eliminating tangential acceleration, but curvature still requires inward radial acceleration.

If 4 m/s is the centre-of-mass speed at h=1 m, a_CM=v²/h=16 m/s² upward. A point on the same pivot-to-centre line at the far tip is 2 m from the pivot, so ω=4 rad/s and a_tip=ω²(2)=32 m/s² upward.

Why: If the speed belongs to the center, divide by its pivot radius for ω, then use each point’s own radius for radial acceleration.

If the diagram’s 4 m/s instead labels that far tip, ω=4/2=2 rad/s, a_tip=8 m/s² and a_CM=4 m/s², both upward. The source wording does not distinguish these interpretations.

Why: If the labeled speed belongs to the far tip, its larger radius gives a different ω and different center acceleration.

As a conditional consistency test, take a uniform solid sphere rigidly pivoted at its surface. The parallel-axis theorem gives I=(2/5)mR²+mh²=(7/5)m·m². Energy then gives ½Iω_bottom²=mgh(1−cos25°).

Why: For the stated solid-sphere test, shift the center inertia to the surface pivot and balance spin energy against gravitational height change.

For this solid-sphere model, v_CM=h√[2mgh(1−cos25°)/I]=1.14529 m/s and the far-tip speed is 2.29058 m/s. Neither is 4 m/s. Conversely, a 4 m/s centre speed requires a turning angle of 98.213°, or a 4 m/s tip speed requires 44.415°, not 25°.

Why: The energy-derived speeds disagree with 4 m/s; solving backward for angle shows the given speed and 25° cannot coexist in that model.

Travel distance is path length ∫|v_tip(t)|dt, not bottom speed multiplied by 5 s. Its value requires a consistent moment of inertia, definition of the tip, and initial phase/time interval. Those are absent or contradictory here, so no unique 5 s distance can be justified.

Why: Oscillating speed changes during the interval, so path length requires its time history and phase, not a constant-speed multiplication.

The period determines ω=2π/T=π/5 rad/s. Choosing maximum deformation at t=0 fixes the temporal factor as cos(πt/5).

Why: The given period fixes temporal frequency, and starting at maximum deformation selects a cosine time factor.

For the intended ideal free–free model, both ends are antinodes, so use cos(kx). Three interior nodes mean n=3 and k=nπ/L=3π/2 rad/m.

Why: Free endpoints need zero slope rather than zero displacement; their antinodes make a cosine shape with three interior nodes.

One valid equation is y(x,t)=(1 m)cos[(3π/2 m⁻¹)x]cos[(π/5 s⁻¹)t]. Its nodes are x=1/3, 1 and 5/3 m, and ∂y/∂x=0 at x=0 and 2 m.

Why: Combine the spatial and temporal factors, then verify the listed interior nodes and zero endpoint slopes.

If instead the ends are fixed and the three nodes include both ends, n=2 and k=π rad/m. A valid alternative is y(x,t)=(1 m)sin[(π m⁻¹)x]cos[(π/5 s⁻¹)t], with nodes at 0, 1 and 2 m.

Why: Fixed endpoints instead need zero displacement; three total nodes would give two half-wave segments and a sine spatial factor.

Other temporal phases are possible. The supplied amplitude and period do not specify a unique phase. A truly slack string needs a tension/model to support a transverse wave. Moreover, A k=3π/2 for the 1 m free–free amplitude is not small, so the stated displacement is a formal standing-wave shape rather than a quantitatively justified small-slope string solution.

Why: The chosen time origin sets a phase, but real wave support still needs a tension model and small slopes; the large stated amplitude fails that latter assumption.

For mÿ+bẏ+ky=0, the underdamped solution is y=A₀e^(−βt)cos(ω_dt+φ), where β=b/(2m) and ω_d²=k/m−β².

Why: Viscous damping changes the oscillation frequency, so relate the measured damped frequency to stiffness using the damping-rate term.

Read β=1/3 s⁻¹, ω_d=4 rad/s and m=4 kg from the corrected temporal expression. The phase −2 rad does not affect k.

Why: Read the decay rate and temporal oscillation rate separately; a phase shift changes timing rather than spring stiffness.

Therefore b=2mβ=8/3 kg/s and k=m(ω_d²+β²)=4(16+1/9)=580/9 N/m=64.4444 N/m.

Why: Recover b from the envelope decay, then add β² back to ω_d² before multiplying by mass to obtain k.

After 12 s, A(12)=10e^(−12/3)=10e^(−4)=0.183156 in the same displacement unit as the initial coefficient 10; this is 1.83156% of A₀.

Why: Insert twelve seconds into the exponential envelope to find the remaining amplitude, keeping the unspecified source displacement unit.

The envelope is positive at every finite t. A zero displacement during an oscillation is a crossing, and a zero instantaneous speed at a turning point is not permanent rest. Exact stopping at 12 s is incompatible with this viscous model.

Why: An exponential envelope never reaches zero at finite time; a temporary crossing or turning point is not permanent stopping.

For a small-amplitude simple pendulum T=2π√(L/g). For a physical pendulum of unchanged shape, T=2π√[I/(mgh)] gives the same inverse-square-root dependence on g.

Why: For unchanged pendulum geometry in the small-amplitude model, gravity enters the period through an inverse square root.

Take the ratio so the unchanged geometry cancels: T_M/T_E=√(g_E/g_M).

Why: Divide the Mars and Earth expressions so the shared length or inertia-to-lever-arm factor cancels.

T_M=(10 s)√(9.8/3.71)=16.2527 s.

Why: Scale the measured Earth period by the square root of the gravity ratio to obtain the Martian period.

The lower Martian gravity weakens the restoring torque, increasing the period. The ratio is dimensionless.

Why: Weaker Martian gravity gives a weaker restoring effect, so the longer computed period has the right direction of change.

The angular frequency is ω=√(k/m)=√(50/0.50)=10 rad/s.

Why: The glider’s spring stiffness and mass fix its ideal angular frequency.

Write x(t)=C cosωt+D sinωt. Initial conditions give C=x₀=0.030 m and D=v₀/ω=−0.040 m.

Why: The cosine coefficient sets initial position, while differentiating the sine term shows its coefficient must be v₀/ω.

Thus x(t)=(0.030 m)cos(10t)−(0.040 m)sin(10t). The amplitude is √(C²+D²)=0.050 m.

Why: The two quadrature coefficients determine the amplitude through their squared sum, while keeping the initial velocity negative.

In x=A cos(ωt+φ), A cosφ=0.030 and −A sinφ=−0.040, so φ=atan2(0.040,0.030)=0.927295 rad. The maximum speed is Aω=0.500 m/s.

Why: Match both initial signs to select the correct phase quadrant; the largest sinusoidal velocity magnitude is Aω.

Static force balance gives kΔℓ=mg, with Δℓ=0.075 m.

Why: At rest the upward spring force balances weight; convert the measured stretch to meters before using Hooke’s law.

Therefore k=(0.60)(9.8)/0.075=78.4 N/m.

Why: Divide the known weight by the equilibrium stretch to determine the spring stiffness.

Measure the moving displacement z from the stretched equilibrium. Gravity and the equilibrium spring force cancel, leaving m d²z/dt²=−kz.

Why: Subtract the equilibrium force balance before describing the motion, leaving only the extra spring force proportional to z.

T=2π√(m/k)=2π√(Δℓ/g)=0.549664 s. Gravity shifts the equilibrium; it does not add a second restoring coefficient about it.

Why: Use that restoring coefficient for the period; gravity enters through the equilibrium stretch rather than a second oscillatory force term.

A rightward displacement stretches the left spring and compresses the right; both push or pull left. Thus F=−(120+180)x=−300x N.

Why: Rightward motion makes both springs exert leftward restoring forces, so their stiffnesses add with one restoring sign.

At x=0.080 m, F=−24.0 N. The sign indicates leftward force.

Why: Insert the release displacement into the combined force law; the negative result identifies the force as leftward.

The elastic energy is E=½(120+180)A²=½(300)(0.080)²=0.960 J.

Why: At release the cart is still, so all mechanical energy is stored in the two deformed springs.

At equilibrium this energy is entirely kinetic: v_max=√(2E/m)=√(1.920/0.75)=1.60 m/s.

Why: At equilibrium both springs are unstrained, so the conserved energy becomes kinetic and the cart reaches maximum speed.

The same force F acts through both massless springs, and the total extension is x=F/k₁+F/k₂.

Why: Massless series springs carry one common force, while their individual extensions add to the total extension.

Hence 1/k_eff=1/200+1/300 and k_eff=120 N/m.

Why: Divide total extension by the common force to obtain reciprocal effective stiffness, rather than adding stiffnesses directly.

The period is T=2π√(1.20/120)=0.628319 s.

Why: The series combination acts as one softer spring in the ideal oscillator period formula.

For total extension 0.050 m, F=k_eff x=6.00 N. The extensions are x₁=6/200=0.030 m and x₂=6/300=0.020 m; their sum is 0.050 m.

Why: Find the common force from total extension, then divide by each stiffness to obtain the two extensions and check their sum.

Momentum is conserved during the brief embedding event: v_after=(0.030×60.0)/(0.500+0.030)=3.39623 m/s.

Why: The impact is brief enough to neglect spring impulse, so pellet-block momentum gives their immediate shared speed.

After impact the oscillating mass is M=0.530 kg. Energy is conserved during the later spring motion, not during the inelastic impact.

Why: Embedding adds the pellet to the oscillating mass; distinguish the energy-losing collision from the later frictionless spring motion.

At the maximum extension, ½Mv_after²=½kA². Thus A=Mv_after/√(kM)=1.8/√(212×0.530)=0.169811 m.

Why: At the first turning point, all postimpact kinetic energy is stored in the spring, setting the oscillation amplitude.

The initial pellet kinetic energy is ½(0.030)(60.0)²=54.0 J. The postimpact mechanical energy is ½(0.530)(3.39623)²=3.05660 J.

Why: Compute mechanical energy immediately before and after embedding, using the different moving masses and speeds.

Impact energy loss is 54.0−3.05660=50.9434 J.

Why: Subtract postimpact mechanical energy from the pellet’s initial energy to measure the inelastic impact loss.

The fall in height is Δh=L(1−cos30°). Conservation of energy gives ½mv²=mgΔh.

Why: The exact cosine geometry gives the bob’s vertical drop, so the speed calculation needs no small-angle approximation.

Therefore v=√[2(9.8)(1.20)(1−cos30°)]=1.77513 m/s.

Why: Convert that gravitational energy drop into the bob’s kinetic energy at the bottom.

At the bottom the inward radial direction is upward. Newton’s second law is F_T−mg=mv²/L.

Why: The bottom’s inward direction is up, making tension positive and weight negative in the radial force balance.

Thus F_T=m(g+v²/L)=0.250×9.8(3−2cos30°)=3.10648 N. Tension exceeds the weight because it must provide the centripetal acceleration.

Why: Add the force needed for upward centripetal acceleration to the weight to obtain a tension greater than mg.

The pivot inertia is I=mL²/12+md² by the parallel-axis theorem.

Why: Moving the pivot away from the center adds md² to the rod’s center-axis inertia.

The squared period is T²=(4π²/g)[L²/(12d)+d], for 0<d≤L/2.

Why: Insert the shifted inertia into the small-angle physical-pendulum period and cancel the rod’s mass.

Minimize the bracket: its derivative is 1−L²/(12d²)=0, giving d=L/√12=0.433013 m. The second derivative L²/(6d³)>0 confirms a minimum.

Why: Differentiate the d-dependent expression and check positive curvature to locate its minimum within the rod’s allowed pivot range.

At this d, L²/(12d)=d. Thus T_min=2π√(2d/g)=1.86781 s. The pivot is L/2−d=0.316987 m from the nearer end.

Why: At the minimum the two terms are equal, simplifying the period; convert d into a location measured from the nearest end.

The rotational equation is Iθ̈=−κθ, so ω=√(κ/I) and T=2π√(I/κ).

Why: A torque proportional to and opposite θ gives rotational harmonic motion, with I playing the role of mass.

Solve I=κ(T/2π)²=0.800(1.60/2π)²=0.0518764 kg·m².

Why: Rearrange the torsional period relation to infer rotational inertia from the measured period and wire stiffness.

At θ=+0.120 rad, τ=−(0.800)(0.120)=−0.0960 N·m. The negative torque acts toward zero angular displacement.

Why: The positive twist produces negative restoring torque, so multiply by κ while keeping the restoring sign.

Corresponding maxima are one damped period apart. Their amplitude ratio is e^(βT_d)=0.0120/0.0100=1.20, where β=b/(2m).

Why: Two successive positive maxima occur at the same phase one damped cycle apart, so their ratio measures envelope decay.

Therefore β=ln(1.20)/0.800=0.227902 s⁻¹ and b=2mβ=0.113951 kg/s.

Why: Take the logarithm of the amplitude ratio to isolate β, then use b=2mβ to obtain damping strength.

The damped angular frequency is ω_d=2π/0.800=7.85398 rad/s.

Why: The time between the maxima gives the damped oscillation frequency, rather than the undamped natural frequency.

Recover the undamped stiffness from ω_d²=k/m−β²: k=0.250(7.85398²+0.227902²)=15.4342 N/m.

Why: Add the decay-rate correction β² to the measured ω_d² before recovering the stiffness.

Check that b<2√(mk), so the motion is underdamped as observed.

Why: Compare the recovered damping with the critical value to confirm the oscillatory model fits the observed repeated maxima.

The critical coefficient is b_c=2√(mk)=2√(0.400×25.0)=2√10=6.32456 kg/s.

Why: The critical coefficient is where the characteristic equation loses its oscillatory roots, setting the comparison threshold for all three carts.

The 3.00 kg/s cart has b<b_c, so it is underdamped. Its β=3.00/(2×0.400)=3.75 s⁻¹.

Why: The first coefficient lies below that threshold; divide it by 2m to obtain the decay rate used in its frequency.

Its damped frequency is ω_d=√(k/m−β²)=√(62.5−14.0625)=6.95971 rad/s.

Why: Damping reduces the squared natural frequency by β², giving the frequency of the cart that still oscillates.

The 2√10 kg/s cart is critically damped, and the 9.00 kg/s cart is overdamped. Neither has a repeating oscillation or an oscillation period.

Why: At or above critical damping the return has no repeated cycle, so neither remaining cart has an oscillation period.

The envelope is A(t)=A₀e^(−bt/2m). Convert the detection threshold to 0.000200 m.

Why: The viscous model gives an exponential amplitude envelope; convert the sensor limit to the same meter unit as A₀.

Set 0.000200=0.0500e^(−0.120t/0.600), or e^(−0.200t)=0.00400.

Why: Set the envelope equal to the detection threshold, rather than setting instantaneous displacement to zero.

Take logarithms: t=ln(0.0500/0.000200)/0.200=27.6073 s.

Why: Take logarithms to undo the exponential and solve for the time needed to decay by that amplitude ratio.

The envelope is monotone, so this is its first threshold crossing. It remains nonzero at any finite time; “no motion” is a sensor criterion here.

Why: The declining envelope crosses the threshold only once; the sensor’s cutoff is finite even though mathematical decay continues.

For a fixed-end string, adjacent mode frequencies differ by f₁=v/(2L). Here f₁=320−240=80.0 Hz.

Why: Fixed-end modes are integer multiples of one fundamental, so adjacent resonances reveal the fundamental frequency through their difference.

The resonances are n=240/80=3 and n=320/80=4.

Why: Divide each measured resonance by that spacing to identify its integer mode number.

The traveling-wave speed is v=2Lf₁=2(0.650)(80.0)=104 m/s.

Why: The fundamental fits half a wavelength into the vibrating length, so v=2Lf₁.

For a taut ideal string v=√(F_T/μ), so F_T=μv²=(2.00×10⁻³)(104)²=21.632 N.

Why: Use the ideal string speed relation to infer tension from the measured wave speed and linear mass density.

Free ends require ∂y/∂x=0 at x=0 and L, so use cos(nπx/L). For n=4, k=4π/1.80 rad/m.

Why: Free ends have zero transverse slope, selecting cosine modes with integer kL/π; the fourth nonzero mode fixes n=4.

The angular frequency is ω=vk=90.0×4π/1.80=200π rad/s, giving f=ω/(2π)=100 Hz.

Why: The mode’s permitted k and common propagation speed give angular frequency, then cycles per second.

With maximum deformation at t=0, y(x,t)=(0.00200 m)cos[(4π/1.80 m⁻¹)x]cos[(200π s⁻¹)t].

Why: The cosine time factor makes the entire guide reach its maximum-deformation shape at the chosen starting time.

Nodes satisfy kx=(j+½)π. For j=0,1,2,3 they are x=0.225, 0.675, 1.125 and 1.575 m. Neither endpoint is a node.

Why: The cosine spatial factor vanishes at odd quarter-cycle positions, giving interior nodes while leaving the free endpoints as antinodes.

The fixed end requires sin(kx); the free-end slope is zero, so cos(kL)=0. Allowed wave numbers are k=(2j−1)π/(2L), with j=1,2,3,….

Why: The fixed end requires zero displacement and the free end zero slope; together they permit only odd quarter-wave combinations.

For the third allowed mode j=3, k=5π/(2×1.40) rad/m. This is the fifth harmonic of the fundamental, not the third harmonic.

Why: The third member of that allowed sequence has five quarter-wavelengths, so it is the fifth harmonic rather than the third.

The wavelength is λ=2π/k=4L/5=1.12 m, and f=v/λ=56.0/1.12=50.0 Hz.

Why: Convert the allowed wave number to wavelength, then divide wave speed by that wavelength for frequency.

Nodes are x=0, λ/2=0.560 m and λ=1.120 m. Antinodes are λ/4=0.280 m, 3λ/4=0.840 m and 5λ/4=1.400 m.

Why: Alternate nodes and antinodes by quarter-wavelength steps, starting with the fixed-end node and ending with the free-end antinode.

Factor out sin(kx). Expanding the shifted cosine gives the temporal factor [3+4cos(2π/3)]cosωt−[4sin(2π/3)]sinωt, in millimetres.

Why: The contributions share one spatial mode; expanding the phase-shifted cosine separates the common temporal cosine and sine parts.

This is (1.00 mm)cosωt−(2√3 mm)sinωt.

Why: Evaluate the 120° phase components, retaining the negative sine coefficient from the cosine addition identity.

Match R cosφ=1.00 and R sinφ=2√3. Then R=√(1+12)=√13 mm=3.60555 mm and φ=atan2(2√3,1)=1.28976 rad.

Why: Match those two components to R cos(ωt+φ); their squared sum gives amplitude and atan2 selects the correct phase.

The spatial nodes stay the same. The amplitudes add to 7.00 mm only when both contributions are in phase.

Why: A shared spatial factor preserves node locations, but out-of-phase temporal components cannot have their amplitudes simply added.

The two terms are the first and second normal modes: k₂=2k₁ and ω₂=2ω₁, consistent with the common wave speed 2 m/s.

Why: Each term has its own integer mode and frequency, with the same ω/k confirming the common propagation speed.

For y to vanish for every t, the coefficients of the different time frequencies must each vanish. Thus sin(πx)=0 and sin(2πx)=0 together.

Why: Different temporal frequencies cannot cancel for every time, so each mode’s spatial coefficient must vanish at a permanent node.

Within 0≤x≤1.00 m, the common permanent nodes are only x=0 and x=1.00 m.

Why: Find the points in the string where both spatial sine factors vanish; only the two endpoints meet that test.

At x=0.50 m the second term is always zero, while the first is 0.010cos(2πt) m. At t=0.25 s that also happens to be zero.

Why: At the midpoint the second mode vanishes always, but the first reaches zero only at the stated instant.

At t=0 the midpoint displacement is +0.010 m, so its later zero is an instantaneous crossing, not a permanent node.

Why: The midpoint moves at other times, which distinguishes that temporary zero from a permanent standing-wave node.

The fundamental has k=π/L and ω=vk=30π/1.20=25π rad/s. Its displacement can be written y=A sin(kx)cosωt.

Why: The fixed-end fundamental sets k=π/L, and the wave speed links it to ω for the chosen mode.

At a time when y=0 everywhere, every element has transverse speed magnitude Aω|sin(kx)|, so all modal energy is kinetic.

Why: At the flat crossing all the energy is kinetic, but each string element has a different speed set by its sine-shaped amplitude.

Integrate: E=½∫₀ᴸ μA²ω²sin²(kx)dx. For an integer mode ∫₀ᴸsin²(kx)dx=L/2.

Why: Add the kinetic energy of all elements using their mass μdx; the sine-squared average over an integer mode is one-half.

Therefore E=μLA²ω²/4=(0.00800)(1.20)(0.00500)²(25π)²/4=0.000370110 J.

Why: Use that integral result with the antinode amplitude and angular frequency to calculate total mode energy.

At maximum deformation, integrating elastic energy ½F_T(∂y/∂x)² gives the same result because F_T=μv² and ω=vk.

Why: At maximum deformation the energy is elastic instead; the string tension-speed relation makes its integral agree with the kinetic calculation.

For linear viscous damping, the steady displacement amplitude is A(Ω)=F₀/√[(k−mΩ²)²+(bΩ)²], and its lag satisfies tanδ=bΩ/(k−mΩ²), using the correct quadrant.

Why: The steady response combines stiffness, inertial and damping terms; its phase must use the correct quadrant when the stiffness-inertia difference changes sign.

At the undamped natural frequency Ω=√(40.0/0.200)=√200=14.1421 rad/s, the stiffness and inertial terms cancel.

Why: At the undamped natural frequency the spring and inertia contributions cancel, leaving damping to set the response.

Then A=F₀/(bΩ)=0.500/(0.800√200)=0.0441942 m and δ=π/2 rad. The response is x=A cos(Ωt−π/2).

Why: With only the damping term left in the denominator, the response has amplitude F₀/(bΩ) and lags the drive by a quarter cycle.

To maximize A, minimize its squared denominator. Differentiation yields Ω_peak²=k/m−b²/(2m²)=200−8=192 s⁻².

Why: Maximizing amplitude means minimizing its squared denominator; differentiating that expression includes the frequency-dependent damping term.

Thus Ω_peak=√192=13.8564 rad/s. With damping, the peak displacement occurs below the undamped natural frequency.

Why: Take the positive square root for the drive frequency; this displacement maximum is shifted below the undamped natural frequency.

In the elevator frame the downward effective gravity is g_eff=g+a for upward elevator acceleration a.

Why: In the upward-accelerating elevator frame, the inertial contribution adds to downward gravity, strengthening the pendulum’s restoring field.

The small-angle relation is T=2π√[L/(g+a)]. Rearranging gives g+a=4π²L/T².

Why: The small-angle period measures that effective field, so rearrange the period equation for g+a.

Hence a=4π²(0.800)/(1.60)²−9.8=2.53701 m/s².

Why: Subtract Earth’s gravity from the inferred effective gravity to isolate the elevator acceleration.

The positive result confirms upward acceleration. A stronger effective downward field shortens the period.

Why: The positive result agrees with upward acceleration and the shorter period expected from stronger effective gravity.

Let θ=8.00t. The condition is |cosθ|≤1/2.

Why: Use the oscillator’s phase as the variable so the displacement-bound condition becomes a cosine inequality.

Within 0≤θ<2π, the allowed intervals are [π/3,2π/3] and [4π/3,5π/3]. Each has width π/3.

Why: Locate both portions of one full phase cycle where the cosine magnitude is at most one-half.

The total allowed phase width is 2π/3, so the time fraction is (2π/3)/(2π)=1/3.

Why: Phase advances uniformly in time, so the fraction of allowed phase equals the fraction of elapsed period.

The period is T=2π/8.00=π/4 s. The total time in the region is T/3=π/12 s=0.261799 s. Endpoints have zero measure and do not change the fraction.

Why: Convert angular frequency to a period and multiply by that fraction to find the total time near equilibrium.