IBDP Physics- C.1 Simple harmonic motion- IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
Time period \( T = 2\pi \sqrt{\frac{m}{k}} \)
For one spring: \( T_1 = 2\pi \sqrt{\frac{m}{k}} \)
For identical springs in parallel: \( k_{eff} = 2k \)
So \( T_2 = 2\pi \sqrt{\frac{m}{2k}} = \frac{T_1}{\sqrt{2}} \)
Therefore \( \frac{T_2}{T_1} = \frac{1}{\sqrt{2}} \)
✅ Answer: (B)
▶️ Answer/Explanation
The frequency of a simple pendulum is given by \( f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}} \).
If the length is halved, \( L \rightarrow \dfrac{L}{2} \), the new frequency is
\( f’ = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L/2}} = \dfrac{1}{2\pi}\sqrt{\dfrac{2g}{L}} \).
Comparing with the original frequency,
\( \dfrac{f’}{f} = \sqrt{2} \).
Hence, the new frequency is \( \sqrt{2}f \).
✅ Answer: (B)
▶️ Answer/Explanation
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)