Unit 7: worked solutions

Use slope = −k/m = −64 s⁻² and solve k = (64)(0.50) = 32 N/m.

Why this step: The signed acceleration-displacement graph measures the restoring acceleration per displacement.

At x = −0.025 m, calculate a = (−64)(−0.025) = +1.60 m/s².

Why this step: The opposite-sign force points toward equilibrium, so the acceleration is rightward.

The cart moving left slows down before turning.

Why this step: Velocity and acceleration have opposite signs; positive acceleration does not mean the cart is already moving right.

Count 20 successive half-cycle trips, in alternating directions, as 10 full cycles.

Why this step: Each trip covers half of the repeated motion; the return trip completes the cycle.

Calculate T = 8.0/10 = 0.80 s and f = 1/T = 1.25 Hz.

Why this step: Period is elapsed time per full cycle, and frequency is its reciprocal.

Using half the correct T would give k_wrong/k_true = (T_true/T_wrong)² = 4.

Why this step: Rearranging T² = 4π²m/k shows why the period-counting error becomes a fourfold spring-constant error.

Read A = 0.050 m and T = 0.40 s, giving ω = 2π/T.

Why this step: Amplitude is the distance from midline to a peak, not the full peak-to-peak distance.

Calculate v_max = Aω = 2π(0.050)/0.40 = 0.785 m/s.

Why this step: The steepest position slope gives maximum speed at equilibrium.

Use a_max = 4π²A/T²: halving T multiplies a_max by 4, from 12.3 to 49.3 m/s².

Why this step: The frequency factor is squared in acceleration. A second position curve has equally high peaks but peaks twice as close in time.

Set K = U = E/2, where E = ½kA².

Why this step: Equal energy shares must each be half of the fixed total.

Solve ½kx² = ¼kA² to get x = ±A/√2.

Why this step: Both sides of equilibrium can have the same potential energy because it depends on x².

Each position is crossed in both travel directions, giving four moments per cycle; speed is v_max/√2.

Why this step: The same energy state does not specify velocity direction. K = E/2 makes squared speed half its maximum.

Plot the three points on axes spanning about s = 0.15 to 0.25 m and F = −1.5 to +1.5 N; label both quantities, scales, and units. A straight line through the points crosses F = 0 at s₀ = 0.200 m.

Why this step: Zero spring force locates equilibrium. The sensor’s coordinate zero does not have to be the spring’s equilibrium.

Find the signed slope: ΔF/Δs = (−1.20 − 1.20)/(0.240 − 0.160) = −30 N/m, giving k = 30 N/m.

Why this step: For an ideal spring, the signed force-displacement slope is −k. Translating the coordinate origin changes the intercept, not the slope.

Define equilibrium displacement x = s − 0.200 m. Then F = −(30 N/m)(s − 0.200 m) and a = F/m = −(100 s⁻²)(s − 0.200 m).

Why this step: The restoring law is proportional to displacement from equilibrium; Newton’s second law converts the net horizontal spring force into acceleration.

At s = 0.230 m, x = +0.030 m and a = −100(0.030) = −3.00 m/s². A cart moving right slows down at that instant.

Why this step: Acceleration points left toward equilibrium and therefore opposes the stated rightward velocity. Using s itself as equilibrium displacement would wrongly give −23 m/s².

Write T = 2π√(m/k).

Why this step: The ideal spring oscillator connects period to mass and stiffness.

Hold k fixed and take T₂/T₁ = √(m₂/m₁) = √9.

Why this step: A ratio cancels constants and avoids unnecessary recalculation.

The period becomes 3 times larger.

Why this step: The square-root mass dependence gives 3, not 9; frequency becomes one-third as large.

Use a horizontal time axis from 0.00 to 0.80 s, such as ticks every 0.10 s, and a vertical displacement axis from −0.10 to +0.10 m, such as ticks every 0.02 m. Plot all nine measured points and draw a smooth sinusoidal trend near them.

Why this step: A numerical plot needs usable scales and units. The measurements have uncertainty, so the physical trend need not pass exactly through every point.

Estimate A ≈ [0.081 − (−0.079)]/2 = 0.080 m. The successive positive peaks near 0.00 and 0.80 s give T ≈ 0.80 s and f ≈ 1.25 Hz.

Why this step: Amplitude is half the peak-to-peak displacement. A period compares repeated positions with the same direction of subsequent motion, not adjacent opposite turning points.

Calculate ω = 2π/T ≈ 7.85 rad/s and v_max = Aω ≈ 0.628 m/s.

Why this step: The greatest slope of the ideal sinusoidal position curve occurs at equilibrium and equals its maximum speed.

At about 0.20 s the curve crosses equilibrium moving left: velocity is most negative and acceleration is approximately zero. At about 0.40 s it reaches the negative turning point: velocity is zero and acceleration is most positive, about +4.93 m/s².

Why this step: Velocity follows position slope, while a = −ω²x gives the restoring acceleration. The displacement minimum has a horizontal tangent but does not have zero acceleration.

An ideal trend x_model = (0.080 m)cos(2πt/0.80 s) differs from the listed positions by at most about 0.001 m, within the stated ±0.002 m uncertainty. The record is consistent with SHM but does not by itself prove the model uniquely; more cycles and a force-displacement test would add evidence.

Why this step: Model checking compares predicted features with measurement uncertainty. Finite noisy measurements support a model without establishing every possible future behavior.

Use total energy E = ½kA² and U = ½kx².

Why this step: These equations describe the horizontal ideal spring system.

At x = A/2, U/E = (x/A)² = 1/4.

Why this step: Potential energy depends on squared displacement, not displacement itself.

Use K = E − U, giving K/E = 3/4.

Why this step: The remaining conserved mechanical energy is kinetic.

Locate the equilibrium position x = 0.

Why this step: The coordinate is measured from the point of zero net restoring force.

Use F = −kx to get F = 0 there.

Why this step: Zero displacement makes the restoring term vanish.

Allow a nonzero velocity through that point.

Why this step: Newton’s second law sets acceleration, not the instantaneous velocity; momentum carries the cart through.

Start from T = 2π√(m/k).

Why this step: We need a graph that turns this square-root dependence into a straight line.

Square both sides: T² = (4π²/k)m.

Why this step: This makes the vertical quantity directly proportional to the horizontal one.

Plot T² against m.

Why this step: Its slope is 4π²/k with units s²/kg, so the graph also lets us determine k.

At the positive turning point, x = +A.

Why this step: The cart has reached its greatest positive displacement.

Read the horizontal tangent of x(t): v = 0.

Why this step: Velocity is the slope of the position graph.

Use a = −ω²x: acceleration is most negative.

Why this step: The greatest positive displacement creates the greatest leftward restoring acceleration.

Write E = ½kA².

Why this step: At a turning point all the oscillator’s energy is potential.

Replace A with 2A: E₂ = ½k(2A)² = 4E₁.

Why this step: Squaring the amplitude doubles the factor twice.

Conclude that energy quadruples.

Why this step: The unchanged mass and stiffness also mean the period stays the same.

Keep signs for both force and displacement.

Why this step: A signed graph must encode the direction of the restoring force.

Use F = −kx with k positive.

Why this step: The force is proportional to displacement and points the opposite way.

Choose a straight line through zero with negative slope.

Why this step: The slope is −k; a positive slope would push the cart away from equilibrium.

Write T = 2π√(L/g).

Why this step: The small-angle pendulum model assumes fixed length and gravitational field.

Notice that m is absent.

Why this step: The mass factor in gravity’s torque cancels the mass factor in inertia.

The period stays unchanged when bob mass doubles.

Why this step: The stated comparison changes none of the variables that set the ideal period.

Keep m and k fixed, so ω = √(k/m) is unchanged.

Why this step: These variables set the ideal frequency and period.

Use v_max = Aω to compare amplitudes.

Why this step: Maximum speed is directly proportional to A at fixed ω.

Doubling A doubles maximum speed and leaves period fixed.

Why this step: The graph stretches vertically while the time between repeated states stays the same.

Distinguish natural-length extension y from equilibrium displacement x = y − mg/k.

Why this step: Gravity has already stretched the spring at equilibrium.

Include U = ½ky² − mgy for spring plus gravity.

Why this step: The chosen spring–mass–Earth system includes both potential contributions.

Shift the potential zero to write U_relative = ½kx².

Why this step: The equilibrium-centered expression represents their combined change, not spring energy alone.

Identify gravity’s torque about the pendulum pivot as −mgL sinθ.

Why this step: The tangential gravity component tends to return the bob to the bottom.

For small θ in radians, replace sinθ by θ.

Why this step: This approximation gives a linear relationship over the small-angle range.

Conclude that restoring torque is proportional to angular displacement.

Why this step: The linear, opposite-sign restoring relation is the requirement for angular SHM.

Use T = 2π√(L/g) for a point bob and small angles.

Why this step: Mass cancels between inertia and the gravitational restoring force.

Doubling bob mass leaves L and g unchanged.

Why this step: It therefore does not change the period under the stated ideal assumptions.

Read a negative slope of position versus time as negative velocity.

Why this step: The slope shows the direction of travel, so the cart is moving left.

At the midline, displacement from equilibrium is zero.

Why this step: The sensor midline represents the coordinate used in the restoring equation.

Use a = −ω²x to get a = 0.

Why this step: Zero acceleration at that instant does not imply zero velocity.

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

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

Use the simple-pendulum scaling T ∝ √L with g fixed.

Why this step: The model assumes small-angle motion and the same gravity.

Multiply the length by 4 to multiply the period by √4 = 2.

Why this step: A linear factor of four in length does not make a factor of four in time.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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