Simple Harmonic Motion 1 Question 1
1. A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K _1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K _2$. Then
(2019 Main, 11 Jan III)
(a) $K _2=2 K _1$
(b) $K _2=\frac{K _1}{2}$
(c) $K _2=\frac{K _1}{4}$
(d) $K _2=K _1$
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Answer:
Correct Answer: 1. (b)
Solution:
- Kinetic energy of a pendulum is maximum at its mean position. Also, maximum kinetic energy of pendulum
$$ K _{\max }=\frac{1}{2} m \omega^{2} a^{2} $$
where,angular frequency
$$ \begin{aligned} \omega & =\frac{2 \pi}{T}=\frac{2 \pi}{2 \pi \sqrt{\frac{l}{g}}} \\ \text { or } \quad \omega & =\sqrt{\frac{g}{l}} \text { or } \omega^{2}=\frac{g}{l} \end{aligned} $$
and $a=$ amplitude.
As amplitude is same in both cases so;
or
$$ \begin{aligned} & K _{\max } \propto \omega^{2} \\ & K _{\max } \propto \frac{1}{l} \end{aligned} $$
$[\because g$ is constant $]$
According to given data, $K _1 \propto \frac{1}{l}$
$$ \begin{array}{ll} \text { and } & K _2 \propto \frac{1}{2 l} \\ \therefore & \frac{K _1}{K _2}=\frac{1 / l}{1 / 2 l}=2 \\ \text { or } & K _1=2 K _2 \Rightarrow K _2=\frac{K _1}{2} \end{array} $$
