Question

A body of unknown mass is attached to an ideal spring with force constant 120 N/m. It is found to vibrate with a frequency of 6.00 Hz.

Find (a) the period; (b) the angular frequency; (c) the mass of the body.

Answer 1

Concepts and reason

The concept used to solve this problem is angular frequency of spring system and period of vibration.

The period of vibration is calculated by taking the inverse of frequency value. Use the expression of angular frequency to find the angular frequency. Use the expression of angular frequency for spring system, re-arrange it for mass value, substitute the value and find the result.

Fundamentals

The period of the vibration of spring system is given as:

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

Here, f is the frequency of vibration.

The expression for the angular frequency of vibration is given as:

$\omega = 2\pi f$

The angular frequency for the spring system is given as follows:

$\omega = \sqrt {\frac{k}{m}}$

Squaring both the sides of the above expression.

$\begin{array}{c}\\{\omega ^2} = \frac{k}{m}\\\\m = \frac{k}{{{\omega ^2}}}\\\end{array}$

(a)

Substitute $6.0{\rm{ Hz}}$for f in equation $T = \frac{1}{f}$.

$\begin{array}{c}\\T = \frac{1}{{6.0{\rm{ Hz}}}}\\\\ = 0.167{\rm{ s}}\\\end{array}$

(b)

Substitute 6.0 Hz for f in $\omega = 2\pi f$.

$\begin{array}{c}\\\omega = 2\pi \left( {6.0{\rm{ Hz}}} \right)\\\\ = 37.68{\rm{ rad/s}}\\\end{array}$

(c)

Substitute 120 N/m for k and $37.68{\rm{ rad/s}}$ for $\omega$in equation $m = \frac{k}{{{\omega ^2}}}$.

$\begin{array}{c}\\m = \frac{{120{\rm{ N/m}}}}{{{{\left( {37.68{\rm{ rad/s}}} \right)}^2}}}\\\\ = 0.084{\rm{ kg}}\\\end{array}$

Ans: Part a

The period of vibration is 0.167 s.

Part b

The angular frequency of the vibration is $37.68{\rm{ rad/s}}$.

Part c

The mass of the body is 0.084 kg.