First slide

Simple hormonic motion

Question

The period of a simple pendulum measured inside a stationary lift is ‘T’. If the lift starts moving upwards with a acceleration g / 3. What will be the time period

Easy

A

T / 3
B

3 T
C

$\large \frac{{\sqrt 3 T}}{2}$
D

$\large \sqrt {\frac{3}{2}} T$

Solution

$\large T\, \propto \frac{1}{{\sqrt g }}$ $\large \Rightarrow \,T\sqrt g \, = \,\operatorname{co} ns\tan t$
g_e = Effective acceleration due to gravity in the lift
$\large = \,g + \frac{g}{3}$ $\large = \frac{{4g}}{3}$
$\large \therefore \,{T^1}\, = \,2\pi \sqrt {\frac{l}{{{g_e}}}}$
$\large \therefore \,{T^1}\sqrt {{g_e}} \, = \,T\sqrt g$
$\large \Rightarrow \,{T^1}\sqrt {\frac{{4g}}{3}} \, = \,T\sqrt g$
$\large \Rightarrow \,{T^1}\, = \,\frac{{\sqrt 3 T}}{2}$

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