(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:

T = 2π ($$\sqrt{\dfrac{m}{k}}$$)

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 2π($$\sqrt{\dfrac{I}{g}}$$). 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 the 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?

Asked by Pragya Singh | 1 year ago |  94

##### Solution :-

(a)  In the case of a simple pendulum, the spring constant k is proportional to the mass. The m is the numerator and the denominator will cancel each other. Therefore, the time period of the simple pendulum is independent of the mass of the bob.

(b) The restoring force acting on the bob of a simple pendulum is given by the expression

F=−mgsinθ

F is the restoring force

m is the mass of the bob

g is the acceleration due to gravity

θ is the angle of displacement

When θ is small, sinθ≈θ.

Then the expression for the time period of a simple pendulum is given by T=2π($$\sqrt{\dfrac{I}{g}}$$)

When θ is large, sinθ<θ. Therefore, the above equation will not be valid. There will be an increase in the time period T.

(c) Wristwatch works on spring action and does not depend on the acceleration due to gravity. Therefore, the watch will show the correct time.

(d) During the free fall of the cabin, the acceleration due to gravity will be zero. Therefore the frequency of oscillation of the simple pendulum will also be zero.

Answered by Pragya Singh | 1 year ago

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