Unit 7: Oscillations
Weightage : 10-15%
A \(0.50~\mathrm{kg}\) object is attached to a vertical spring of spring constant \(k\), as shown above. The object is pulled downward and released. The object oscillates vertically. If upward is the positive direction, the position \(x\) of the object as a function of time \(t\) is given by
\(x=\beta\sin(\omega t+\phi)\),
where \(\beta=0.20~\mathrm{m}\), \(\omega=4.0~\mathrm{rad\,s^{-1}}\), and \(\phi=\dfrac{\pi}{3}~\mathrm{rad}\).
The period of oscillation of the object is most nearly
Correct Answer: \( \boxed{\mathrm{D}} \)
For simple harmonic motion, the period depends only on the angular frequency:
\(T=\dfrac{2\pi}{\omega}\)
Substituting \(\omega=4.0~\mathrm{rad\,s^{-1}}\),
\(T=\dfrac{2\pi}{4.0}=\dfrac{\pi}{2}\approx1.57~\mathrm{s}\)
The amplitude \(\beta\) and phase constant \(\phi\) affect the motion’s displacement at a given instant but do not affect the period.
Therefore, the period of oscillation is approximately \(1.6~\mathrm{s}\), so the correct answer is (D).
A mass \(M\) suspended from a spring with spring constant \(k\) has a period \(T\) when set into simple harmonic motion on Earth.
What is the period of oscillation on Mars, whose mass is approximately \(\dfrac{1}{9}\) that of Earth and whose radius is approximately \(\dfrac{1}{2}\) that of Earth?
Correct Answer: \( \boxed{\mathrm{C}} \)
The period of a mass-spring oscillator is
\(T=2\pi\sqrt{\dfrac{M}{k}}\)
Notice that the expression depends only on the mass \(M\) and the spring constant \(k\). It does not depend on the acceleration due to gravity.
Although the gravitational acceleration on Mars is different from that on Earth, gravity only changes the equilibrium position of the spring. It does not change the frequency or period of the oscillation.
Therefore,
\(T_{\mathrm{Mars}}=T_{\mathrm{Earth}}=T\)
Hence, the correct answer is (C).
A mass \(m\) attached to a spring oscillates on a horizontal surface with period \(T\). The total mechanical energy of the oscillation is \(E\). Suppose a new mass \(4m\) oscillates on the same spring with the same amplitude. What are the new period and the total mechanical energy?
| Option | Period | Total Mechanical Energy |
|---|---|---|
| (A) | \(T\) | \(E\) |
| (B) | \(2T\) | \(E\) |
| (C) | \(2T\) | \(2E\) |
| (D) | \(T\) | \(4E\) |
| (E) | \(2T\) | \(16E\) |
Correct Answer: \( \boxed{\mathrm{B}} \)
The period of a spring-mass oscillator is
\(T=2\pi\sqrt{\dfrac{m}{k}}\)
Replacing the mass \(m\) with \(4m\),
\(T’=2\pi\sqrt{\dfrac{4m}{k}}=2\left(2\pi\sqrt{\dfrac{m}{k}}\right)=2T\)
The total mechanical energy of a spring-mass oscillator is
\(E=\dfrac{1}{2}kA^2\)
Since the spring constant \(k\) and the amplitude \(A\) remain unchanged, the total mechanical energy is unchanged and does not depend on the mass.
Therefore,
\(T’=2T,\qquad E’=E\)
Hence, the correct answer is (B).
