A student investigates simple harmonic motion with a mass-spring system that oscillates horizontally on a frictionless surface. In a first trial, the spring is compressed and the mass is released from rest at time \(t=0\). The mass oscillates with a period \(T\).
(a) State the energy changes that take place:
(i) between \(t=0\) and \(t=\dfrac{T}{4}\).
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(ii) between \(t=\dfrac{T}{4}\) and \(t=\dfrac{T}{2}\).
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(b) In a second trial, the mass is tripled without changing the spring.
Determine the ratio
\(\dfrac{\text{frequency of oscillation in the first trial}}{\text{frequency of oscillation in the second trial}}\)
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(c) Describe what property the spring must have for the motion of the system to be simple harmonic.
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Most-appropriate topic codes (IB Physics 2025):
(a)(i) Correct Answer:
At \(t=0\), the spring is compressed and the mass is released from rest. Therefore, the system initially has maximum elastic potential energy and zero kinetic energy.
As the mass moves from \(t=0\) to \(t=\dfrac{T}{4}\), the elastic potential energy decreases and is converted into kinetic energy.
Thus, \( \boxed{\text{elastic/spring potential energy} \rightarrow \text{kinetic energy}} \).
(a)(ii) Correct Answer:
At \(t=\dfrac{T}{4}\), the mass passes through the equilibrium position, where its speed and kinetic energy are maximum.
From \(t=\dfrac{T}{4}\) to \(t=\dfrac{T}{2}\), the mass moves towards the opposite maximum displacement. Its speed decreases to zero while the spring becomes increasingly compressed or stretched.
Therefore, kinetic energy is converted back into elastic potential energy.
Thus, \( \boxed{\text{kinetic energy} \rightarrow \text{elastic/spring potential energy}} \).
(b) Correct Answer: \( \boxed{\sqrt{3}} \)
For a mass-spring system, the frequency is
\(f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\)
In the second trial, the mass is tripled:
\(m_2=3m_1\)
Therefore,
\(f_2=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{3m_1}}\)
Hence,
\(\dfrac{f_1}{f_2}=\sqrt{3}\)
Therefore, the required ratio is \( \boxed{\sqrt{3}} \), approximately \(1.7\).
(c) Correct Answer:
The spring must obey Hooke’s law, so that the force exerted by the spring is directly proportional to the displacement from equilibrium.
This can be written as
\(F=-kx\)
where \(k\) is constant. The negative sign indicates that the restoring force acts opposite to the displacement.
