Question

The position of a 0.30 kg object attached to a spring is described by x = (0.25 m) cos(0.4 π t).

(a) What is the amplitude of the motion?

(b) Calculate the spring constant.

(c) Calculate the position of the object at t = 0.30 s.

(d) Calculate the velocity of the object at t = 0.30 s.

Answer 1

A.

Standard SHM equation is given by:

x = X0*cos wt

X0 = Amplitude

from given equation

X0 = 0.25 m = Amplitude

B.

w = sqrt (k/m)

k = m*w^2

k = 0.30*(0.4*pi)^2 = 0.47 N/m

C.

when t = 0.30 sec

x = 0.25*cos (0.4*pi*t)

x = 0.25*cos (0.4*pi*0.3)

x = 0.25*cos (0.12*pi)

x = 0.23 m

D.

v = dx/dt

v = d(0.25*cos (0.4*pi*t))/dt

v = -0.25*0.4*pi*sin (0.4*pi*t)

v = -0.1*pi*sin (0.4*pi*t)

at t = 0.30 sec

v = -0.1*pi*sin (0.4*pi*0.3)

v = -0.1*pi*sin (0.12*pi)

v = -0.12 m/sec