Home / IBDP Physics- C.1 Simple harmonic motion- IB Style Questions For SL Paper 1A

IBDP Physics- C.1 Simple harmonic motion- IB Style Questions For SL Paper 1A -FA 2025

Question

A mass attached to a spring undergoes simple harmonic motion with a time period \(T_{1}\).
A second identical spring is connected in parallel to the first spring, as illustrated. The mass remains unchanged.
The combined system oscillates with a time period \(T_{2}\). What is the ratio \(\frac{T_{2}}{T_{1}}\)?
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{\sqrt{2}}\)
(C) \(\sqrt{2}\)
(D) 2
▶️ Answer/Explanation
Detailed solution

1. Period of a Spring–Mass System:
\(T = 2\pi\sqrt{\frac{m}{k}}\).

2. Effect of Parallel Springs:
For two identical springs connected in parallel, the effective spring constant is \(k_{\text{eff}} = k + k = 2k\).

3. Compare the Time Periods:

  • \(T_{1} = 2\pi\sqrt{\frac{m}{k}}\)
  • \(T_{2} = 2\pi\sqrt{\frac{m}{2k}} = \frac{1}{\sqrt{2}}T_{1}\)

4. Ratio of Periods:
\(\frac{T_{2}}{T_{1}} = \frac{1}{\sqrt{2}}\).
Answer: (B)

Question

A mass is oscillating with simple harmonic motion. At time \(t\), the acceleration is a positive maximum. What are the displacement and velocity of the mass at time \(t\)?
OptionDisplacementVelocity
Apositive maximumzero
Bnegative maximumzero
Cpositive maximumnegative maximum
Dnegative maximumnegative maximum
▶️Answer/Explanation
In SHM, \(a=-\omega^2 x\).
For \(a\) to be a positive maximum, \(x\) must be a negative maximum (most negative displacement).
At maximum displacement (an extreme position), the mass momentarily turns around, so \(v=0\).
Answer: (B)

Question

Which graph shows the variation with time \(t\) of the kinetic energy (KE) of an object undergoing simple harmonic motion (SHM) of period \(T\)?
▶️ Answer/Explanation
Detailed solution

In SHM, the velocity varies sinusoidally, for example \(v = v_{\max}\sin(\omega t + \phi)\).
The kinetic energy is \( \mathrm{KE} = \tfrac12 mv^2 \), so
\( \mathrm{KE} \propto \sin^2(\omega t + \phi) \).

Since \(\sin^2(\cdot)\) is always \(\ge 0\), the KE graph never goes negative.
Also, \(\sin^2(\omega t)\) has period \(\tfrac{\pi}{\omega} = \tfrac{T}{2}\), so over the time interval \(0 \to 2T\) there are \( \dfrac{2T}{T/2} = 4 \) identical “humps”.

Therefore the correct graph is the one that stays above zero and repeats every \(T/2\): Graph D.

Answer: (D)

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