Period and Frequency: Worked Solutions

Try the questions on the lesson page first. Then compare both the answer and the reasoning. The worked steps here are original.

Question 1: A spring oscillator mass increases by a factor of 9; its spring is unchanged. What happens to T?

A spring oscillator mass increases by a factor of 9; its spring is unchanged. What happens to T?

1. Write T = 2π√(m/k). Why: The ideal spring oscillator connects period to mass and stiffness.
2. Hold k fixed and take T₂/T₁ = √(m₂/m₁) = √9. Why: A ratio cancels constants and avoids unnecessary recalculation.
3. The period becomes 3 times larger. Why: The square-root mass dependence gives 3, not 9; frequency becomes one-third as large.

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Question 2: Which graph gives a straight line for the ideal spring model?

Which graph gives a straight line for the ideal spring model?

1. Start from T = 2π√(m/k). Why: We need a graph that turns this square-root dependence into a straight line.
2. Square both sides: T² = (4π²/k)m. Why: This makes the vertical quantity directly proportional to the horizontal one.
3. Plot T² against m. Why: Its slope is 4π²/k with units s²/kg, so the graph also lets us determine k.

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Question 3: A small-angle pendulum bob mass doubles at fixed L and g. Its period…

A small-angle pendulum bob mass doubles at fixed L and g. Its period…

1. Write T = 2π√(L/g). Why: The small-angle pendulum model assumes fixed length and gravitational field.
2. Notice that m is absent. Why: The mass factor in gravity’s torque cancels the mass factor in inertia.
3. The period stays unchanged when bob mass doubles. Why: The stated comparison changes none of the variables that set the ideal period.

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Question 4: an experiment with the wrong cycle count

A student times 20 successive turning-point-to-turning-point trips of a spring oscillator, counting both directions: right to left, then left to right, and so on. The total time is 8.0 s. The student treats each trip as a full cycle and reports T = 0.40 s. Correct the cycle count, find f, and explain the effect on a spring constant found from this T.

1. Count 20 successive half-cycle trips, in alternating directions, as 10 full cycles. Why: Each trip covers half of the repeated motion; the return trip completes the cycle.
2. Calculate T = 8.0/10 = 0.80 s and f = 1/T = 1.25 Hz. Why: Period is elapsed time per full cycle, and frequency is its reciprocal.
3. Using half the correct T would give k_wrong/k_true = (T_true/T_wrong)² = 4. Why: Rearranging T² = 4π²m/k shows why the period-counting error becomes a fourfold spring-constant error.

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