Question

A body executes simple harmonic motion according to following equation.X=5.0cos[2πt+π4] m where t=1.5 s, calculate 1 displacement 2 velocity 3 acceleration

Answer 1

Answer:

x = -5/√2 m.

v = 10π/√2= 22.21 m/s

a= 20π^2/√2= 139.58 m/s^2

Explanation:

1. displacement at t=1.5=3/2 is

x=5 cos[3π+π/4] = -5/√2m.

2. velocity (v) = dx/dt= -10π sin[2πt+π/4]

at t=1.5=3/2,

v = 10π/√2= 22.21 m/s

acceleration (a) = dv/dt = -20π^2 cos[2πt+π/4]

at t=1.5=3/2,

a= 20π^2/√2= 139.58 m/s^2