A `100g` block is connected to a horizontal massless spring of force constant `25.6 N//m`. The block is free to oscillate on a horizontal fricationless surface. The block is displced by `3 cm` from the equilibrium position, and at `t = 0`, it si released from rest at `x = 0`, The position-time graph of motion of the block is shown in figure.

Let us now make a slight change to the initial conditions. At `t = 0`, let the block be released from the same position with an initial velocity `v_(1) = 64 cm//s`. Position of the block as a function of time can be expressed as

Options

A. `x = 5 sin 16t`
B. `x = 5sin(16t+37^(@))`
C. `x = 5sin(16t-37^(@))`
D. `x=5cos(16t +37^(@))`

Answer

B

Explanation

Let `x = A sin(16t + phi) v = Aomega cos (16t + phi)` where `omega^(2) = (K)/(m) = (25.6)/(0.1) = 256 rArr omega = 16"rad"/"sec"`
At`t = 0, 3 = A sinphi` & `64 = Acosphi rArr tanphi = 3/4 rArr phi = 37^(@)` Also `A = 5 cm`
Therefore equation of SHM `x = 5 sin (16t + 37^(@))`