Find the amplitude, period, and frequency of the resulting Simple Harmonic Motion.

Question

A body of mass 5 kg is suspended by a spring, which stretches 0.1 m when the body is attached. It is then displaced downward as additional 0.05m and released. Find the amplitude, period, and frequency of the resulting Simple Harmonic Motion.

Answer 1

The first thing to do is to find the spring constant, k.   Hook's law is used to do this:  F = -k x.   Take the positive

direction of displacements to be downward.  So the spring exerts an upward force

of magnitude  k 0.1  This balances the downward force of gravity the

magnitude of which is  5 g   (g = 9.81).    Setting these magnitudes equal (the force balance)

gives  k =  5 (9.81)/0.1  =  490.5  N/m.

 

Next is the analysis of the SHM.    the angular frequency, ω ,  is given by ω =sqrt(k/m)

This evaluates to 9.90 radian/s.    The linear frequency, f,  is ω/ (2 π) =  1.57 cycles/s.

The period, T,  is  T = 1/f =  0.634 s.

 

The amplitude is simply the additional downward displacement = 0.05 m.