Question

At t=0 , observer and source are at same place. Now the source is projected with velocity 60 sqrt2 m/s at 45 ^(@) . Natural frequency of source is 1000 Hz.find the frequency heard by the observer at t=2s. Take speed of sound = 340 m/s

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

the frequency heard by the observer at t=2s is $823.52 hz$

Explanation:

Doppler Effect is the change in frequency when the position of the observer changes with respect to the source. However, here we are assuming that the velocity of the wave is constant during the interaction. Also, the wave is either approaching the observer or moving away from the observer, only.

Formula used:

$f=\left(\frac{c \pm v_{o}}{c \pm v_{s}}\right) f_{0} ,$ where, $f$is the apparent frequency of the sound, $f_{0}$ is the actual or real frequency of the sound wave, and C  is the speed of the sound wave, and $v_{s}$, $v_{o}$ is the speed of the moving source and observer respectively.

Let us consider the given sound wave to travel at a speed $c=60 \mathrm{~m} / \mathrm{s} .$

Given that the source is moving towards the stationary observer at a velocity of $60 \mathrm{~m} / \mathrm{s}$, then the speed of the observer is $v_{o}=0 \mathrm{~m} / \mathrm{s}$, and the

speed of the source with respect to the observer is $v_{s}=50 \mathrm{~m} / \mathrm{s} ,$ as the source is moving towards the observer.

Also given that the apparent or observed frequency of the sound wave is $f=1000 \mathrm{~Hz} .$

From Doppler's law we know $f=\left(\frac{c \pm v_{o}}{c \pm v_{s}}\right) f_{0}$

Here, the Doppler's law will become $f=\left(\frac{c}{c-v_{s}}\right) f_{0}$, as the source is moving towards the stationary observer .

Now substituting the values, we get

$1000=\left(\frac{340}{340-60}\right) f_{0}$

$1000 =(\frac{340}{280}) f_{0} \\\frac{1000*280}{340}=f_{0}\\ f_{0} =823.52\ hz$