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

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 :

$f=\dfrac{1}{T}$

$f=\dfrac{1}{0.507}$

f = 1.97 Hz

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

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

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

$k=4\pi^2\times 0.324\times 1.97^2$  

k = 49.64 N/m

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

$E=\dfrac{1}{2}m\omega^2A^2$

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

$A=\sqrt{\dfrac{E}{2\pi^2f^2m}}$

$A=\sqrt{\dfrac{0.263}{2\pi^2\times (1.97)^2\times 0.324}}$

A = 0.102 meters

Hence, this is the required solution.