Question

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is 2 s.

Answer 1

Answer:

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is 2 s. After another 2 s it returns to B. The ratio of distance OB to its amplitude is 1: √2. As T = 8s, OB = xB = A. Sin( (2* 3.14)/T). 1 s. or, OB: A = A/ √2 : A. = 1: √2.

Answer 2

Explanation:

$\huge \tt A : B \: = 1 : \sqrt{2} \\ \\ \large \tt \: \: \: \: \: (or) \\ \\ \huge \tt \frac{ \: \: A \: \: }{ \: \: B \: \: } \: \: = \: \: \frac{1}{ \: \: \sqrt{2} \: \: }$