Question

An observer sitting in between two vertical walls claps 10 times per second. He is 17 m away from one of the wall and his first clap coincides with the echo then Calculate the velocity of sound.

please answer it's really Urgent. Please help. ​

Answer 1

$\rm\red{\underline{\underline{✧ Question :- }}}$

• Q.An observer sitting in between two vertical walls claps 10 times per second. He is 17 m away from one of the wall and his first clap coincides with the echo then Calculate the velocity of sound ??

$\rm\blue{\underline{\underline{✧ Given :- }}}$

• The frequency of the claps is 10Hz.
• The distance taken here is 2 × 17 = 34metre. ( as it is reflected back or echo ).

$\rm\orange{\underline{\underline{✧ To \: find :- }}}$

• The velocity of the sound = ??

$\rm\green{\underline{\underline{✧ <u>Let's</u> <u>\</u><u>:</u><u>Solve</u> :- }}}$

$\bf{ We \: know \: that \: time \: taken \: is} \\ \bf { \: the \: inverse \: of \: frequency.} \\ \bf{ \therefore} {The \: time \: taken \: is : - \: } \\ { \therefore\{t \: = \frac{1}{f} = \frac{1}{10} = 0.1 \: seconds} \\ \bf \:\purple {Now, \: velocity \: or\: speed \: of \:the \: sound : - } \\ \: =\blue{ \frac{total \: \: distance \: \: travelled(m)}\orange{total \: \: time \: \: taken (s)} } \\ \tt \: = \frac{34 \: m}{0.1 \: s}= 340m/s \\ \bf{ \therefore}the \: speed \: of \: the \: sound \: is \: 340m/s.$

Answer 2

Answer:

\rm\red{\underline{\underline{✧ Question :- }}}

✧Question:−

Q.An observer sitting in between two vertical walls claps 10 times per second. He is 17 m away from one of the wall and his first clap coincides with the echo then Calculate the velocity of sound ??

\rm\blue{\underline{\underline{✧ Given :- }}}

✧Given:−

The frequency of the claps is 10Hz.

The distance taken here is 2 × 17 = 34metre. ( as it is reflected back or echo ).

\rm\orange{\underline{\underline{✧ To \: find :- }}}

✧Tofind:−

The velocity of the sound = ??

\rm\green{\underline{\underline{✧ Let's \:Solve :- }}}

✧Let

′

sSolve:−

\begin{gathered}\bf{ We \: know \: that \: time \: taken \: is} \\ \bf { \: the \: inverse \: of \: frequency.} \\ \bf{ \therefore} {The \: time \: taken \: is : - \: } \\ { \therefore\{t \: = \frac{1}{f} = \frac{1}{10} = 0.1 \: seconds} \\ \bf \:\purple {Now, \: velocity \: or\: speed \: of \:the \: sound : - } \\ \: =\blue{ \frac{total \: \: distance \: \: travelled(m)}\orange{total \: \: time \: \: taken (s)} } \\ \tt \: = \frac{34 \: m}{0.1 \: s}= 340m/s \\ \bf{ \therefore}the \: speed \: of \: the \: sound \: is \: 340m/s.\end{gathered}

Weknowthattimetakenis

theinverseoffrequency.

∴Thetimetakenis:−

∴{t=

f

1

=

10

1

=0.1seconds

Now,velocityorspeedofthesound:−

=

totaltimetaken(s)

totaldistancetravelled(m)

=

0.1s

34m

=340m/s

∴thespeedofthesoundis340m/s.