### Q.16:- Answer the following questions:

*k*and mass

*m*of the particle:

*T*= 2π √

*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 √*l*/*g*

### (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 cabinthat is freely falling under gravity?

**Answer:-**(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

*F* = –*mg* sin*θ*

where,

*F *= Restoring force

*m* = Mass of the bob

*g* = Acceleration due to gravity

*θ* = Angle of displacement

For small* θ*, sin*θ ≈ **θ*

For large* θ*, sin*θ* is greater than* θ*.

This decreases the effective value of *g*.

Hence, the time period increases as:

*T* = 2π √*l*/*g*

*
*(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.

(d) Gravity disappears for a man under free fall, so frequency is zero.