Question

The period of small vertical oscillations of a spring pendulum
is found to be equal to 2 s. The whole apparatus is now placed
in an elevator accelerating upwards with an acceleration 0.5 g.
The percentage change in the period of the oscillations will be
4%
2%
3%
0%

Answer 1

Given:

The period of small vertical oscillations of a spring pendulum is found to be equal to 2 s. The whole apparatus is now placed in an elevator accelerating upwards with an acceleration 0.5g.

To find:

% change in time period ?

Calculation:

Let's find out the time period of the oscillations in stationary elevator.

At equilibrium:

$\therefore \: mg = kx_{0}$

When displaced by small distance:

$F = mg - k(x + x_{0})$

$\implies F = kx_{0} - k(x + x_{0})$

$\implies F = - kx$

$\implies a = - \dfrac{kx}{m}$

$\implies - { \omega}^{2}x = - \dfrac{kx}{m}$

$\implies \: \omega = \sqrt{ \dfrac{k}{m} }$

$\implies \: T = \dfrac{2\pi}{\omega}$

$\implies \: T = 2\pi \sqrt{ \dfrac{m}{k} }$

$\implies \: T \propto {g}^{0}$

So, here gravitational force and it's changes (due to acceleration of lift) doesn't make any change in TIME PERIOD.

% change in T will be 0% .