Question

A bolt of mass 2.08×10?2 kg moves with simple harmonic motion that has an amplitude of 0.243 m and a period of 1.480 s. The displacement of the bolt is 0.243 m when t = 0.
(a) Find the displacement of the bolt at time t = 0.508 s.

(b) Find the magnitude of the force acting on the bolt at time t = 0.508 s.
(c) What is the minimum time required for the bolt to move from its initial position to the point x = -0.176 m?
(d) What is the speed of the bolt at x = -0.176 m?
(e) Plot the displacement, velocity and acceleration graphs.

Answer 1

The displacement of the body at time t executing SHM is given by

x = A cos(2 ? f t + ?)

where, A = amplitude, f = frequency,  ? = phase in radians

the period , time taken by one cycle is

T = 1/f

given that T = 1.480

hence f = 1/1.480 = 0.676 cycles/s

also, m = 0.0208 kg, A = 0.243

x = 0.243 m when t = 0

implies that 0.243 = 0.243 cos?

so ? = 0

a) we have

x = 0.243 cos(2 ? f t)

at t = 0.508

x = 0.243*cos(2*3.14*0.676*0.508) = 0.243*cos(2.157 radians) = 0.243*(-0.553) = -0.134 m.

b) F = -kx

we know that

T = 1/f = 2 ? sqrt[m/k]

1.480 = 2 ? sqrt[0.0208/k]

implies that k = 0.374 N/m

so, F = -k*x = -(0.374)(-0.134) = 0.050 N , the force is acting in positive direction.

c) for x = -0.176, we get

-0.176 = 0.243 cos[2*3.14*0.676*t]

cos[4.24*t] = -0.724

4.24*t = 2.38 radians

hence t= 2.38/4.24 = 0.561 s

d) speed is given by

V = - A*(2 ? f) *sin[2 ? f t]

at x = 0.176 m

we get

V = -(0.243*2*3.14*0.676)*sin[2*3.14*0.676*0.561] = -1.03*sin[2.377] = -0.71 m/s

so |V| = 0.71 m/s.