First slide

Applications of SHM

Question

The time period of a simple pendulum inside a stationary lift is $\sqrt{5}$ sec. What will be the time period when the lift moves upward with an acceleration g/4?

Moderate

A

$\sqrt{5}$ seconds
B

$2 \sqrt{5}$ seconds
C

$(2 + \sqrt{5})$ seconds
D

2 seconds

Solution

$time period of a simple pendulum is T=2π \sqrt{\frac{l}{g}} ----(1)and time period of a simple pendulum i n a l i f t a c c e l e r a t i n g u p w a r d i s T^{1} = 2π \sqrt{\frac{l}{(g+a)}} ⇒ T^{1} = 2π \sqrt{\frac{l}{(g+g/4)}} {⇒T}^{1} = 2π \sqrt{\frac{l}{(5g/4)}} ---(2) d i v i d e e q u a t i o n (1) b y (2) \Rightarrow \frac{T}{T^{1}} = \sqrt{\frac{5}{4}} \Rightarrow T^{1} = \frac{2 T}{\sqrt{5}}, b y q u e s t i o n T = \sqrt{5}, s u b s t i t u t e i n a b o v e e q u a t i o n \Rightarrow T^{1} = 2 s e c o n d s$

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