physics numericals for class 9 sound with answers

Time period for 20 Hz = 1/f= 1/20=0.05s
Time period for 20,000 Hz = 1/f=1/20000=$5 \times 10^{-5}$ s
Hence, the time period range is 0.05 s to 5 x 10^-5 s.

Speed= distance/time
or
$v= \frac {\lambda}{T}$
or
$v= \lambda \nu$

$v= \lambda \nu$
$v= 1.10 \times 312=343.2$ m/s

$v= \lambda \nu$
$ \lambda= \frac {v}{{\nu}$
$\lambda= \frac {340}{512}=.664$ m

Given,
Speed of wave, v = $3 \times 10^8$ m/s
Time taken for receiving the radar signal
t =$2 \times 10^{-5}$ sec
Distance travelled by the wave
= $v \times t = 3 \times 10^8 \times 2 \times 10^{-5}=6 \times 10^3$ m
In this time wave has to travel twice the distance between the aeroplane and receiver
Hence, the distance from aeroplane
= $3 \times 10^3$ m

$\nu =2000 $ hz
$\lambda = .4$ m
Speed of the sound
$v= \lambda \nu= 2000 \times .4=800$ m/s
Time taken to cover 1.6 km = 1600/800=2 sec

$v= \lambda \nu$
$\nu= \frac {v}{\lambda}$
As frequency is constant
$\frac {v_{1}}{\lambda _{1}}=\frac {v_{2}}{\lambda _{2}}$
$\lambda _{2}= \frac {v_{2} \times \lambda _{1}}{v_{1}}= \frac { 3v \times \lambda}{v}=3 \lambda $ m

The time interval between successive compressions from the source is equal to the time period and time period is reciprocal of the frequency. Therefore, it can be calculated as follows:
T= 1/F
T= 1/500
T = 0.002 s.

Distance of wall from the boy=85m
To hear the echo, sound has to travel a distance =2 * 185=170 m
Since 2 claps are produced in one second, each clap is produced after =1/2=.5 sec
Since echo matches with the second clapping.
Speed of sound= 170/.5=340 m/s

Here we need to first find the time for the stone to drop in pond
$s=ut + \frac {1}{2}gt^2$
Here s=500m, u=0,g=10m/s²
$500=\frac {1}{2} \times 10 \times t^2$
or t=10 sec
Now time for the sound to travel to peak of tower
=500/340=1.47 sec
So total time to heard the sound=10 + 1.47=11.47 sec

Speed of light is greater than the speed of sound. Therefore, lightning will reach in no time but the thunder will be heard 3 seconds after the lightning by the man.
We know that the lightning and the sound travels the same distance, let us consider that as 'd'.
Now for light
$d= 3 \times 10^8 t$
For sound
$d=330 \times (t+3)$ as sound is heard 3 sec later
Equating both
$3 \times 10^8 t= 330(t+3)$
Solving for t,we get
$t=3.30 \times 10^{-6}$ s
Therefore
$d=3 \times 10^8 \times 3.30 \times 10^{-6}=990$ m

The wavelength of a wave is the distance between two consecutive crests or troughs. Here, the distance between a crest and next trough is half of the wavelength.
$\frac {\lambda}{2}=10$ cm
$\lambda$=20 cm =.2 m
Frequency is the number of oscillations per unit time.
f=10 hz
Now, velocity of the wave,
$v= \lambda \nu = .2 \times 10=2 $ m/s

$v= \lambda \nu$
$\nu= \frac {v}{\lambda}$
As frequency is constant
$\frac {v_{air}}{\lambda _{air}}=\frac {v_{hel}}{\lambda _{hel}}$
$\lambda _{hel}= \frac {v_{hel} \times \lambda _{air}}{v_{air}}= \frac { 970 \times .3}{340}=.85$ m

Let v be the speed of the engine
Initial distance of engine from cliff=900m
Distance of engine from cliff when echo is heard= 900 -5v
Total distance covered by sound= 900 + 900 -5v=1800-5v
Now
Speed= distance/time
340=1800-5v/5
or
v=30 m/s

Given
Time in which echo returned, t = 6 s,
Speed of sound, v = 342 m/s
Distance = Speed × Time = 342 × 6 = 2052m
As this distance is twice the distance of reflecting surface from source.
So,
The distance of reflecting surface from source = 2052 / 2 = 1026 m.