Question

10) A 500 gram mass vibrates according to the equation x = 0.45cos6.5t.

a) What is the amplitude of vibration?

b) What is the period of the vibration?

c) What is the frequency of vibration?

d) What is the total energy of the vibration (where k = mω2 )?

e) What is the potential energy when it is at x= 0.22 m

Answer 1

The general form of simple harmonic motion is expressed as:

x = A cos (t + )

Here, A = amplitude

   = angular frequency

= initial phase

Comparing the given equation in question with the general form of the equation.

(a) Amplitude, A = 0.45 m

(b) Period of vibration, t = 1/

Angular frequency, = 6.5 rad/s

2 =

2 3.14 = 6.5 rad/s

= 1.035 Hz

t = 0.96 s

(c) Frequency of vibration, = 1.03 Hz

(d) Energy of vibration, E = 1/2 mA2

E = 1/2 ( 0.5 kg)(6.5 rad/s)2(0.45 m)2

E = 2.13 J

(e) Potential energy at x = 0.22 m is:

PE = 1/2 mx2

PE = 1/2 ( 0.5 kg)(6.5 rad/s)2(0.22 m )2

PE = 0.511 J