First slide

Applications of SHM

Question

A man measure the period of a simple pendulum inside a stationary lift and finds it to be T seconds. If the lift accelerates upwards with an accelerating g/4, then the period of the pendulum will be

Moderate

A

$T$
B

$\frac{T}{4}$
C

$\frac{2T}{\sqrt{5}}$
D

$2 T \sqrt{5}$

Solution

in stationary lift $T=2π \sqrt{\frac{l}{g}}$ ------(1)
In upward motion of lift
$T^{1} =2π \sqrt{\frac{l}{g+a}}$ -----(2) here a is acceleration of lift
divide equation (2) by (1)

$\frac{T^{1}}{T} = \sqrt{\frac{g}{g+a}} = \sqrt{\frac{g}{g+g/4}}$ given a=g/4
$\to \frac{T^{1}}{T} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$
cross multiply, to obtain new time period of simple pendulum
$\Rightarrow T^{1} = \frac{2T}{\sqrt{5}}$ .

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