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Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant $k$ and mass $m$ of the particle.
$T=2 \pi \sqrt{\frac{m}{k}}$. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$. Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
(a) Time period of a particle in SHM depends on the force constant $k$ and mass $m$ of the particle.
$T=2 \pi \sqrt{\frac{m}{k}}$. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$. Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Solution:
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Verified Answer
(a) In case of a spring, $k$ does not depend upon $m$. However, in case of a simple pendulum, $k$ is directly proportional to $m$ and hence the ratio $\frac{m}{k}$ is a constant quantity.
(b) The restoring force for the bob of the pendulum is given by
If $\theta$ is small, then $\sin \theta=\theta=\frac{y}{l} \quad \therefore F=-\frac{m g}{l} y$ i.e., the motion is simple harmonic and time period is $T=2 \pi \sqrt{\frac{l}{g}}$.
Clearly, the above formula is obtained only if we apply the approximation $\sin \theta \approx \theta$. For large angles, this approximation is not valid and $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$.
(c) The wristwatch uses an electronic system or spring system to give the time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.
(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.
(b) The restoring force for the bob of the pendulum is given by
If $\theta$ is small, then $\sin \theta=\theta=\frac{y}{l} \quad \therefore F=-\frac{m g}{l} y$ i.e., the motion is simple harmonic and time period is $T=2 \pi \sqrt{\frac{l}{g}}$.
Clearly, the above formula is obtained only if we apply the approximation $\sin \theta \approx \theta$. For large angles, this approximation is not valid and $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$.
(c) The wristwatch uses an electronic system or spring system to give the time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.
(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.
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