JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 10)
Assume that the displacement(s) of air is
proportional to the pressure difference ($$\Delta $$p)
created by a sound wave. Displacement (s)
further depends on the speed of sound (v),
density of air ($$\rho $$) and the frequency (f). If
$$\Delta $$p ~ 10 Pa, v ~ 300 m/s, $$\rho $$ ~ 1 kg/m3 and f ~ 1000 Hz,
then s will be of the order of (take the
multiplicative constant to be 1) :
1 mm
$${3 \over {100}}$$ mm
10 mm
$${1 \over {10}}$$ mm
Explanation
Given, S $$ \propto $$ $$\Delta $$p
and Proportionally constant = 1
We know,
$$\Delta $$p = S$$\beta $$k
= $$\rho $$v2 $$ \times $$ $${\omega \over v} \times S$$
= $${\rho v\omega S}$$
$$ \therefore $$ S = $${{\Delta p} \over {\rho v\omega }}$$
= $${{\Delta p} \over {\rho v2\pi f}}$$
= $${{\Delta p} \over {\rho vf}}$$
[As Proportionally constant = 1 so assume 2$$\pi $$ = 1]
$$ = {{10} \over {1 \times 300 \times 1000}}$$
$$ = {1 \over {30}}mm$$
$$ \approx {3 \over {100}}mm$$
and Proportionally constant = 1
We know,
$$\Delta $$p = S$$\beta $$k
= $$\rho $$v2 $$ \times $$ $${\omega \over v} \times S$$
= $${\rho v\omega S}$$
$$ \therefore $$ S = $${{\Delta p} \over {\rho v\omega }}$$
= $${{\Delta p} \over {\rho v2\pi f}}$$
= $${{\Delta p} \over {\rho vf}}$$
[As Proportionally constant = 1 so assume 2$$\pi $$ = 1]
$$ = {{10} \over {1 \times 300 \times 1000}}$$
$$ = {1 \over {30}}mm$$
$$ \approx {3 \over {100}}mm$$
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