First slide

Applications of SHM

Question

A simple pendulum has time period ' $T_{1}$ '. The point of suspension is now moved upwards according to the relation $y = K T^{2} . (K = 1 m / s e c^{2})$ where y is the vertical displacement. The time period now becomes ' $T_{2}$ '. Then find the ratio of $\frac{{T_{1}}^{2}}{{T_{2}}^{2}}$ $(g = 10 m / s e c^{2})$

Moderate

A

$\frac{5}{6}$
B

$\frac{6}{5}$
C

$\frac{4}{3}$
D

$\frac{3}{4}$

Solution

$v e r t i c a l d i s p l a c e m e n t y = u t + \frac{1}{2} a t^{2} h e r e u = 0, h e n c e, y = \frac{1}{2} a t^{2} g i v e n y = k t^{2} c o m p a r i n g, {y=kt}^{2} = \frac{1}{2} {at}^{2} \Rightarrow k= \frac{1}{2} \times a s u b s t i t u t i n g k = 1 \Rightarrow 1= \frac{1}{2} \times a \Rightarrow {a=2m/sec}^{2} T_{1} = 2π \sqrt{\frac{l}{g}} T_{2} = 2π \sqrt{\frac{l}{g+a}} o n d i v i d i n g a b o v e t w o e q u a t i o n s, a n d s u b s t i t u t e a = 2 \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{g+a}{g} = \frac{10 + 2}{10} = 6 / 5$

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