First slide

Applications of SHM

Question

Calculate the change in the length of a simple pendulum of length 1m, when its period of oscillation changes from 2 sec to 1.5 sec

Moderate

A

$\frac{- 7}{16} m$
B

$\frac{7}{16} m$
C

$\frac{16}{7} m$
D

$\frac{- 16}{7} m$

Solution

$T i m e p e r i o d = T = 2 π \sqrt{\frac{l}{g}} where l = length of pendulum g = acceleration due to gravity T α \sqrt{l} \frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}} b y q u e s t i o n T_{1} = 2, T_{2} = 1.5 \frac{2}{1.5} = \sqrt{\frac{l_{1}}{l_{2}}}, s q u a r i n g o n b o t h s i d e s \Rightarrow l_{2} = \frac{9}{16} \Rightarrow Δ l = l_{2} - l_{1} Δ l = \frac{9}{16} - 1 c h a n g e i n l e n g t h = Δ l = \frac{- 7}{16}$

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