Unit 7: Oscillations
Weightage : 10-15%
A pendulum is released from an initial angle of \(45^\circ\). Which statement is most accurate?
Correct Answer: \(\boxed{\mathrm{B}}\)
The expression
\(T=2\pi\sqrt{\dfrac{L}{g}}\)
is valid only for small oscillations where \(\sin\theta\approx\theta\). At an initial angle of \(45^\circ\), the approximation is no longer exact, so the actual period is slightly larger.
A pendulum clock keeps accurate time on Earth. The clock is taken to a planet where the gravitational acceleration is \(\dfrac{g}{4}\). Without adjusting the pendulum, the clock will
Correct Answer: \(\boxed{\mathrm{C}}\)
Since
\(T\propto\dfrac{1}{\sqrt{g}}\),
reducing \(g\) by a factor of 4 doubles the period, so the clock runs slow.
A uniform rod of length \(L\) is pivoted about one end and oscillates with small amplitude as a physical pendulum. Another uniform rod has the same mass but length \(2L\), also pivoted about one end.
The ratio of the periods of the second rod to the first rod is
Correct Answer: \(\boxed{\mathrm{B}}\)
For a uniform rod pivoted about one end,
\(I=\dfrac13mL^2,\qquad d=\dfrac{L}{2}\)
Therefore,
\(T=2\pi\sqrt{\dfrac{I}{mgd}} =2\pi\sqrt{\dfrac{\frac13mL^2}{mg(L/2)}} =2\pi\sqrt{\dfrac{2L}{3g}}\)
Hence,
\(T\propto\sqrt{L}\)
Doubling the rod length increases the period by a factor of \(\sqrt2\). Therefore, the correct answer is (B).
