Question

A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity v = Kt where $K=4.4m{s}^{-2}$, the new time period will be (Take $g=10m{s}^{-2}$)

• A. $\frac{9}{4}s$
• B. $\frac{5}{3}s$
• C. 2.5 s
• D. 4.4 s

Answer 1

The correct option is C

2.5 s

Original time period is

$T_1=2\pi \sqrt{\frac{l}{g}}\dots \left(1\right)$

When the pendulum is moving upwards, the effective value of g is
$g_{eff}=g+a$

where a is the acceleration of the pendulum which is given by

$a=\frac{dv}{dt}=\frac{1}{dt}\left(Kt\right)=K=4.4m{s}^{-2}$

$\therefore g_{eff}=g+a=10+4.4=14.4m{s}^{-2}$

Therefore, the new time period is

$T_2=2\pi \sqrt{\frac{1}{g_{eff}}}$

$\frac{T_2}{T_1}=\sqrt{\frac{g}{g_{eff}}}=\sqrt{\frac{10}{14.4}}=\frac{1}{1.2}$

Hence the correct choice is (c).