Question

A mass of 0.380 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by

x(t) = (0.800 m)cos[(10.0 rad/s)t].

Determine the following.

(a) amplitude of oscillation for the oscillating mass
____m

(b) force constant for the spring
____ N/m

(c) position of the mass after it has been oscillating for one half a period
______ m

(d) position of the mass one-sixth of a period after it has been released
____ m

(e) time it takes the mass to get to the position

x = ?0.300 m after it has been released

____ s

Answer 1

amplitude of oscillation for the oscillating mass is 0.800 m

force constant for the spring is 0.380 kg * (10.0 rad/s)^2 = 38 N/m

period is T= 2*3.14/10= 0.63s

So, position of the mass after it has been oscillating for one half a period is x= (0.800 m)cos[(10.0 rad/s)0.63/2] = -0.800m

position of the mass one-sixth of a period after it has been released is x= (0.800 m)cos[(10.0 rad/s)0.63/6] = 0.400m

time it takes the mass to get to the position x = ?0.300 m after it has been released is found by -0.300 = (0.800 m)cos[(10.0 rad/s)t]. So, time it takes is t= 0.1955s