Question

Two sources of sound are moving in opposite directions with velocities $v_{1}and{v}_2(v_1>v_2)$. Both are moving away from a stationary observer. The frequency of both the sources is $1700Hz$. What is the value of $(v_1-v_2)$ so that the beat frequency observed by the observer is $10Hz$? $v_{sound}=340m/s$ and assume that $v_{1}and{v}_2$ both are very much less than $v_{sound}$ :

• A. $1m/s$
• B. $2m/s$
• C. $3m/s$
• D. $4m/s$

Answer 1

The correct option is B $2m/s$

Velocity of sound $v_{sound}=340m/s$

Velocities of source 1 and source 2 are $v_1$ and $v_2$ respectively.

Frequency of sources $\nu =1700Hz$

Doppler effect when source moves away from the stationary observer :

Apparent frequency heard by the observer ${\nu }^{\prime }=\nu [\frac{v_{sound}}{v_{sound}+v_{source}}]$

Frequency heard by observer due to source 1, ${\nu }_1=1700[\frac{340}{340+v_1}]Hz$

Due to source 2, ${\nu }_2=1700[\frac{340}{340+v_2}]Hz$

Now ${\nu }_2-{\nu }_1=10$

$1700[\frac{340}{340+v_2}]$ $-1700[\frac{340}{340+v_1}]=10$

$1700\left(340\right)$ $[\frac{v_1-v_2}{(340+v_1)(340+v_2)}]=10$

Now $v_1,$ and $v_2$ can be neglected from $340$

$1700\left(340\right)[\frac{v_1-v_2}{(340{)}^2}]=10⟹v_1-v_2=2m/s$