A simple pendulum has time period T1 . The point of suspension is now moved upwards according to the relation y=kt2.(k=1 m/s2) where y is the vertical displacement. The time period now becomes T2 .The ratio of T21T22=x10. Find the value of x (g=10m/s2)

A simple pendulum has time period T1 . The point of suspension is now

1.	A simple pendulum has time period T1 . The point of suspension is now moved upwards according to the relation y=kt2.(k=1 m/s2) where y is the vertical displacement. The time period now becomes T2 .The ratio of T21T22=x10. Find the value of x (g=10m/s2)
Answer» A simple pendulum has time period $T_{1}$ . The point of suspension is now moved upwards according to the relation $y = k t^{2} . (k = 1 m / s^{2})$ where $y$ is the vertical displacement. The time period now becomes $T_{2}$ .The ratio of $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{x}{10}$ . Find the value of x $(g = 10 m / s^{2})$

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A simple pendulum has time period T1 . The point of suspension is now moved upwards according to the relation y=kt2.(k=1 m/s2) where y is the vertical displacement. The time period now becomes T2 .The ratio of T21T22=x10. Find the value of x (g=10m/s2)

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