Question

When a mass of m = 324 g is attached to a spring and the mass-spring system is set into oscillatory motion, the period of the motion is T = 0.507 s. Determine the following.

(a) frequency of the motion in hertz Hz

(b) force constant of the spring N/m

(c) amplitude of the oscillation, if the total energy of the oscillating system is 0.263 J

Answer 1

Step-by-step explanation:

It is given that,

Mass of the object, m = 324 g = 0.324 kg

The period of motion, T = 0.507 s

(a) Let f is the frequency of the motion. It is given by :

`f=(1)/(T)`

`f=(1)/(0.507)`

f = 1.97 Hz

(b) In terms of force constant, the frequency of motion is given by :

`f=(1)/(2\pi)\sqrt{(k)/(m)}`

`k=4\pi^2mf^2`

`k=4\pi^2* 0.324* 1.97^2`

k = 49.64 N/m

(c) The total energy of the oscillating system is given by :

`E=(1)/(2)m\omega^2A^2`

`E=2\pi^2f^2mA^2`

`A=\sqrt{(E)/(2\pi^2f^2m)}`

`A=\sqrt{(0.263)/(2\pi^2* (1.97)^2* 0.324)}`

A = 0.102 meters

Hence, this is the required solution.