A mass-spring system oscillates on a frictionless horizontal surface with simple harmonic motion. The mass \(m\) oscillates with angular frequency \(\omega\).
The total energy is \(E_{\mathrm{T}}=\dfrac{1}{2}m\omega^{2}x_{0}^{2}\), where \(x_{0}\) is the maximum displacement.
(a) Show that \(E_{\mathrm{T}}\) can be written as \(E_{\mathrm{T}}=\dfrac{1}{2}kx_{0}^{2}\), where \(k\) is the spring constant.
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In a first trial, the spring is compressed by \(0.15\,\mathrm{m}\), and the mass is released from rest.
\(k=24.5\,\mathrm{N\,m^{-1}}\)
(b) Calculate the maximum elastic potential energy of this system.
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In a second trial, the mass is doubled, and the spring is compressed by the same amount as before.
(c) Determine
\(\dfrac{\text{maximum speed of the mass in the first trial}}{\text{maximum speed of the mass in the second trial}}\)
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Most-appropriate topic codes (IB Physics 2025):
(a) Correct Answer: \( \boxed{E_{\mathrm{T}}=\dfrac{1}{2}kx_{0}^{2}} \)
For a mass-spring system undergoing simple harmonic motion,
\(\omega^{2}=\dfrac{k}{m}\)
Therefore,
\(m\omega^{2}=k\)
The given total energy is
\(E_{\mathrm{T}}=\dfrac{1}{2}m\omega^{2}x_{0}^{2}\)
Substituting \(m\omega^{2}=k\),
\(E_{\mathrm{T}}=\dfrac{1}{2}kx_{0}^{2}\)
Hence, \( \boxed{E_{\mathrm{T}}=\dfrac{1}{2}kx_{0}^{2}} \).
(b) Correct Answer: \( \boxed{0.28\,\mathrm{J}} \)
At maximum displacement, all the energy is stored as elastic potential energy.
Using
\(E_{\mathrm{T}}=\dfrac{1}{2}kx_{0}^{2}\)
with \(k=24.5\,\mathrm{N\,m^{-1}}\) and \(x_{0}=0.15\,\mathrm{m}\),
\(E_{\mathrm{T}}=\dfrac{1}{2}(24.5)(0.15)^{2}\)
\(E_{\mathrm{T}}=0.2756\,\mathrm{J}\)
Therefore, \( \boxed{E_{\mathrm{T}}\approx0.28\,\mathrm{J}} \).
(c) Correct Answer: \( \boxed{\sqrt{2}} \)
At the equilibrium position, the elastic potential energy is zero and the total energy is kinetic energy:
\(E_{\mathrm{T}}=\dfrac{1}{2}mv_{\max}^{2}\)
Therefore,
\(v_{\max}=\sqrt{\dfrac{2E_{\mathrm{T}}}{m}}\)
The maximum compression is the same in both trials and \(k\) is unchanged, so
\(E_{\mathrm{T}}=\dfrac{1}{2}kx_{0}^{2}\)
is also the same in both trials.
The second trial has twice the mass, so
\(\dfrac{v_{\max,1}}{v_{\max,2}}=\sqrt{\dfrac{m_{2}}{m_{1}}}\)
Since \(m_{2}=2m_{1}\),
\(\dfrac{v_{\max,1}}{v_{\max,2}}=\sqrt{2}\)
Hence, the required ratio is \( \boxed{\sqrt{2}} \).
