Question

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=2πkm​​. 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 the larger angles of oscillation, a more involved analysis shows that T is greater than 2πgl​​. Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch in 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?

Answer 1

Hint:If in a simple pendulum, a bob of mass m start oscillations then time period depends on length of the string and acceleration due to gravity. There is also a restoring force act on the bob and it depends on mass of bob m, acceleration due to gravity and angle of displacement.

Formula used
Restoring force,
F=-mg\[\sin \theta \]
Where, m is mass of the bob, g is acceleration due to gravity and \[\theta \] is angle of displacement.
Time period of a simple pendulum is
\[T = 2\pi \sqrt {\dfrac{l}{g}} \]
Where t is the time period, l is the length of string and g is acceleration due to gravity.
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Where m is mass and k is spring constant.

Complete step by step solution:
(a) In a simple pendulum, force constant or spring factor is proportional to mass as,
\[k \propto m\]
As Time period of simple pendulum is,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Here mass m cancels out in numerator and denominator. Therefore the time period of a simple pendulum is independent of the mass of the bob.

(b) In simple pendulum ,restoring force is given as
F=-mg\[\sin \theta \]
For a small angle, \[\sin \theta \approx \theta \]
And for a large angle, \[\sin \theta > \theta \] which decreases the effective value of g.
Thus we get the time period increases as:
\[T = 2\pi \sqrt {\dfrac{l}{g}} \]

(c) The answer is Yes because a wristwatch does not depend on gravity and only depends on spring action. Hence the watch gives the correct time during the free fall.

(d) If the object is freely falling under gravity, there is no effect of gravity on it. Hence no change in the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity

Note: In a simple pendulum time period is calculated as the time taken by the pendulum to complete one complete oscillation. Also, in simple harmonic motion(SHM) the time period can be determined by one complete oscillation. In a pendulum, the time period is independent of the mass.