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=10m/sec2)

Moderate

A

$\frac{5}{6}$
B

$\frac{6}{5}$
C

$\frac{4}{3}$
D

$\frac{3}{4}$

Solution

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

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