First slide

Applications of SHM

Question

A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the percentage increase in the time period of the pendulum of increased length is

Moderate

A

10%
B

21%
C

30%
D

50%

Solution

If initial length $l_{1} = 100$ then $l_{2} = 121$
By using $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}}$
Hence, $\frac{T_{1}}{T_{2}} = \sqrt{\frac{100}{121}} \Rightarrow T_{2} = 1 .1 T_{1}$
% increase = $\frac{T_{2} - T_{1}}{T_{1}} \times 100 = 10 %$

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