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

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

An object undergoes simple harmonic motion with frequency \(f\). The graph shows the variation of its acceleration with displacement.

What is the value of \(f^2\) in \(\mathrm{s^{-2}}\)?

(A) \(\dfrac{5}{\pi^2}\)
(B) \(\dfrac{5}{\pi}\)
(C) \(\dfrac{100}{\pi^2}\)
(D) \(\dfrac{100}{\pi}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

For simple harmonic motion, the acceleration is related to displacement by

\(a=-\omega^2x\)

Therefore, the gradient of an acceleration-displacement graph is

\(\dfrac{a}{x}=-\omega^2\)

From the graph, two points are approximately \((-0.20\,\mathrm{m},4\,\mathrm{m\,s^{-2}})\) and \((0.20\,\mathrm{m},-4\,\mathrm{m\,s^{-2}})\).

The gradient is

\(\dfrac{-4-4}{0.20-(-0.20)}=-20\,\mathrm{s^{-2}}\)

Hence,

\(\omega^2=20\,\mathrm{s^{-2}}\)

Using

\(\omega=2\pi f\)

we have

\((2\pi f)^2=20\)

\(4\pi^2f^2=20\)

Therefore,

\(f^2=\dfrac{20}{4\pi^2}=\dfrac{5}{\pi^2}\)

Hence, the correct answer is \( \boxed{\mathrm{A}} \).

Question 

A mass oscillating vertically on a spring undergoes simple harmonic motion with amplitude \(X\), total energy \(E\) and maximum speed \(v_{\max}\).

What is correct about the elastic potential energy \(E_H\) and speed \(v\) when the displacement is \(\dfrac{X}{2}\)?

 Elastic potential energySpeed
(A)\(E_H<\dfrac{E}{2}\)\(v>\dfrac{v_{\max}}{2}\)
(B)\(E_H<\dfrac{E}{2}\)\(v<\dfrac{v_{\max}}{2}\)
(C)\(E_H>\dfrac{E}{2}\)\(v>\dfrac{v_{\max}}{2}\)
(D)\(E_H>\dfrac{E}{2}\)\(v<\dfrac{v_{\max}}{2}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

For a mass-spring system, the elastic potential energy is

\(E_H=\dfrac{1}{2}kx^2\)

The total energy is the elastic potential energy at maximum displacement \(X\):

\(E=\dfrac{1}{2}kX^2\)

When \(x=\dfrac{X}{2}\),

\(E_H=\dfrac{1}{2}k\left(\dfrac{X}{2}\right)^2=\dfrac{1}{4}\left(\dfrac{1}{2}kX^2\right)\)

Therefore,

\(E_H=\dfrac{E}{4}<\dfrac{E}{2}\)

The remaining energy is kinetic energy:

\(\dfrac{1}{2}mv^2=E-E_H=\dfrac{3E}{4}\)

At maximum speed,

\(\dfrac{1}{2}mv_{\max}^2=E\)

Thus,

\(\dfrac{v^2}{v_{\max}^2}=\dfrac{3}{4}\)

\(v=\dfrac{\sqrt{3}}{2}v_{\max}\)

Since \(\dfrac{\sqrt{3}}{2}>\dfrac{1}{2}\),

\(v>\dfrac{v_{\max}}{2}\)

Hence, the correct answer is \( \boxed{\mathrm{A}} \).

Question

Which graph represents the variation with displacement of the potential energy \(P\) and the total energy \(T\) of a system undergoing simple harmonic motion (SHM)?
C.1 Simple harmonic motion HL Paper 1
▶️ Answer/Explanation
Detailed solution

In simple harmonic motion, the potential energy varies with the square of the displacement from the equilibrium position, producing a parabolic curve. The total energy of the system remains constant and is represented by a horizontal line.

Among the given graphs, only option (B) correctly shows a parabolic variation of potential energy with displacement and a constant total energy.

✅ Answer: (B)

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