# 1) Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average...

## Question:

1) Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average frequency will you hear, and what will the beat frequency be?

2) Twin jet engines on an airplane are producing an average sound frequency of 4100 Hz with a beat frequency of 0.500 Hz. What are their individual frequencies?

## Beats

Consider two tuning forks emitting harmonic waves with the same amplitude {eq}\displaystyle {a} {/eq} at angular frequencies {eq}\displaystyle {\omega_1} {/eq} and {eq}\displaystyle {\omega_2} {/eq}. Then the individual wavefunctions at the point of observation say x=0, have the form,

{eq}\displaystyle {y_1=a\ sin (\omega_1 t)} {/eq}

And,

{eq}\displaystyle {y_2= a \sin (\omega_2 t)} {/eq}.

The waves superpose to give the resultant effect. Then the net displacement is,

{eq}\displaystyle {y=y_1+y_2=a \sin (\omega_1 t)+a \sin (\omega_2 t)=2a \sin \left(\frac{(\omega_1+\omega_2)t}{2} \right) \cos \left(\frac{(\omega_1-\omega_2)t}{2} \right)} {/eq}.

The expression may be interpreted as the representation of an oscillation of frequency {eq}\displaystyle { \frac{\omega_1+\omega_2}{2}} {/eq} with a time varying amplitude {eq}\displaystyle { 2a\cos \left(\frac{(\omega_1-\omega_2)t}{2} \right)} {/eq} that is in itself oscillating with a frequency {eq}\displaystyle {\frac{\omega_1-\omega_2}{2}} {/eq}. Since the intensity or the energy crossing unit perpendicular area per second is proportional to the square of the amplitude, it will oscillate like the square of the cosine function with a frequency {eq}\displaystyle {|\omega_1-\omega_2|} {/eq}. This is what is directly observed and is called the angular beat frequency.

1)

Here two tuning forks having frequencies 460 Hz and 464 Hz are sounded together. So the average frequency heard is,

{eq}\displaystyle {\nu_{avg}=\frac{\nu_1+\nu_2}{2}=\frac{460+464}{2}=462\ Hz} {/eq}.

The beat frequency will be,

{eq}\displaystyle {\nu_{beat}=\nu_2-\nu_1=464-460=4\ Hz} {/eq}.

2)

It is given that twin jet engines produce sound of average frequency 4100 Hz and beat frequency 0.5 Hz.

That is,

{eq}\displaystyle {\nu_1-\nu_2=0.5}----------(1) {/eq}

And,

{eq}\displaystyle { \frac{\nu_1+\nu_2}{2}=4100} {/eq}

Or,

{eq}\displaystyle {\nu_1+\nu_2=8200\ Hz}--------(2) {/eq}.

Solving (1) and (2), we get,

{eq}\displaystyle { \nu_1=4100.25\ Hz} {/eq}

And,

{eq}\displaystyle { \nu_2=4099.75\ Hz} {/eq}.