# Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when...

## Question:

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 348 m/s, piano A produces a wavelength of 0.766 m, while piano B produces a wavelength of 0.780 m.

How much time separates successive beats?

## Beat Period:

When two waves of slightly different frequencies are superimposed and the total intensity is observed at a fixed point in space, then it will be seen that the intensity varies over time. The intensity itself will oscillate with a characteristic frequency. This phenomenon is called beats and the frequency is called the beat frequency. The beat frequency equals the difference between the source frequencies. Its reciprocal is the beat period. Beats may be regarded as interference in the time domain as opposed to the fringe patterns detected in the space domain.

## Answer and Explanation:

The frequency of a wave {eq}\displaystyle {\nu} {/eq}, the wavelength {eq}\displaystyle {\lambda} {/eq} and the propagation velocity {eq}\displaystyle {v} {/eq} are related according to,

{eq}\displaystyle {v=\nu \lambda} {/eq}.

Hence the frequency,

{eq}\displaystyle { \nu=\frac{v}{\lambda}} {/eq}.

Here it is given that the speed of sound is {eq}\displaystyle {v=348\ m/s} {/eq}.

Piano A produces a wavelength of 0.766 m. Hence,

{eq}\displaystyle { \nu_A=\frac{348}{0.766}=454.3\ Hz} {/eq}.

Similarly for Piano B,

{eq}\displaystyle {\lambda=0.780\ m} {/eq}.

Therefore,

{eq}\displaystyle { \nu_B=\frac{348}{0.78}=446.2\ Hz} {/eq}.

Hence the beat frequency is,

{eq}\displaystyle {\nu_b=\nu_A-\nu_B=454.3-446.2=8.1\ Hz} {/eq}.

That is the time between successive beats or the beat period is,

{eq}\displaystyle { T=\frac{1}{\nu_b}=\frac{1}{8.1}=0.124\ s} {/eq}.

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from MEGA Physics: Practice & Study Guide

Chapter 15 / Lesson 10