First slide

Applications of SHM

Question

If the length of the simple pendulum is increased by 44%, then what is the change in time period of pendulum

Moderate

A

22%
B

20%
C

33%
D

44%

Solution

$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{T_{2}}{T_{1}} = \sqrt{\frac{l_{2}}{l_{1}}} = \sqrt{\frac{144}{100}} = \frac{12}{10}$
$\Rightarrow$ T₂ = 1.2 T₁
Hence % increase $= \frac{T_{2} - T_{1}}{T_{1}} \times 100 = 20 %$

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