Question

A man measures the period of simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates downwards with acceleration of $\frac{g}{4}$, the period of oscillation will be

• A. $T\times \frac{\sqrt{3}}{2}s$
• B. $T\times \frac{2}{\sqrt{3}}s$
• C. $\frac{T}{2}s$
• D. $\sqrt{T}s$

Answer 1

The correct option is B $T\times \frac{2}{\sqrt{3}}s$

Effective acceleration due to gravity when the lift accelerates downwards,
$g^{\prime }=g-\frac{g}{4}=\frac{3g}{4}$
Time period $T=2\pi \sqrt{\left(\frac{l}{g}\right)}$ ...... (i)
$T^{\prime }$ $=2\pi \sqrt{\left(\frac{l}{g^{\prime }}\right)}$ $=2\pi \sqrt{\left(\frac{4l}{3g}\right)}=2\pi \times \frac{2}{\sqrt{3}}\sqrt{\left(\frac{l}{g}\right)}$ ....(ii)
Now dividing Eq. (ii) by Eq. (i), we get
$\frac{T^{\prime }}{T}=\frac{2}{\sqrt{3}}$

or $T^{\prime }=T\times \frac{2}{\sqrt{3}}$