Question

A boy standing in front of a wall at $17m$ produces 10 claps per second. He notices that the sound of his clapping coincides with the echo. Echo is heard only once when clapping is stopped. Calculate the speed of sound:
$340\dfrac{m}{s}$.
$360\dfrac{m}{s}$.
$290\dfrac{m}{s}$.
$300\dfrac{m}{s}$

Answer 1

An echo can be only heard when sound travels from the source to the wall, hits it and comes back to the source; this means that the sound is travelling a distance from the source to the wall twice as it returns to the source. Also the frequency is the number of oscillations per second.

Step by step solution:
It is given that the number of claps per second is 10.
$T = \dfrac{1}{f} \\\ T = \dfrac{1}{{10}} \\\ T = 0 \cdot 1\sec \\\$
So the time taken for 1 clap is 0.1sec.
It is given that the wall is at a distance of 17m away from the source. Also in order to hear echo the sound has to travel twice the distance between source and wall. So the total distance travelled by the sound is given by,
$d = \left( 2 \right) \times \left( {17} \right)$
$d = 34m$
As it is given that the time to travel this distance should be $T = 0.1\sec$ because in order to hear an echo the sound of clap being produced and the sound after reflection should meet.
Now, let us calculate the velocity of the sound, as we know that${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$.
Since the distance is given as $d = 34m$ and also the time taken is $T = 0 \cdot 1\sec$.
So,
${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
${\text{speed}} = \dfrac{{34}}{{0 \cdot 1}}$
${\text{speed}} = {\text{340}}\dfrac{m}{s}$
So, in order to hear the echo the speed of sound should be$v = {\text{340}}\dfrac{m}{s}$.

So the correct answer for this problem is option A.

Note: The students should be clear why we took twice the distance between the wall and the source. In order to hear an echo the sound producing clap should be mixed with the sound of clap coming after reflection from the wall and that is why we take the twice distance between the source and the wall. While solving the problems related to sonar we do similar calculations.