\(S_1\) and \(S_2\) are two identical sound sources of frequency \(656\) \(Hz\). The source \(S_1\) is located at \(π\) and \(π_2\) moves anti-clockwise with a uniform speed \(4\sqrt2\) \(m s ^{β1}\) on a circular path around \(π\), as shown in the figure. There are three points \(π\), \(π\) and \(π
\) on this path such that \(π\) and \(π
\) are diametrically opposite while \(π\) is equidistant from them. A sound detector is placed at point \(π\). The source \(π_1\) can move along direction \(ππ\). [Given: The speed of sound in air is \(324\) \(m s ^{β1}\) ]
When only \(π_2\) is emitting sound and it is at \(π\), the frequency of sound measured by the detector in Hz is _________.
When only \(π_2\) is emitting sound and it is at \(π\), the frequency of sound measured by the detector in Hz is _________.
Correct Answer: 648
Solution and Explanation
When only \(π_2\) is emitting sound and it is at \(π\), the frequency of sound measured by the detector in Hz is \(\underline{648.00}.\)
Explanation:
The Doppler effect formula for sound when the source is moving towards the observer is given by:
\(f'=f\times\frac{v+v_0}{v+v_s}\)ββ
Where:
- \('f\)β² is the observed frequency,
- \(f\) is the emitted frequency,
- \(v\) is the speed of sound in the medium (given as 324 msβ1),
- \(v_o\)β is the speed of the observer (in this case, at point π),
- \(v_s\)β is the speed of the source.
Initially, only π2 emits sound at π. To calculate the frequency observed at π, let's determine the relative speed of π2 towards the detector at π.
The observer's speed \((v_oβ)\) is 0 since the detector is stationary at π.
Now, the speed of π2 concerning the detector at π will be the component of the speed of π2 perpendicular to the line ππ.
Given that π2 moves at a uniform speed of 4\(\sqrt{2}\) \(ms^{β1}\) on a circular path around π, and π is equidistant from π and π
, which are diametrically opposite, the speed of π2 towards the detector at π is 4\(\sqrt{2}\)β \(ms^{β1}\)
Now, let's use the Doppler effect formula:
\(f'=f\times\frac{v+v_0}{v+v_s}\)
Given that \(f\)= \(656\) Hz, \(v\) = \(324 \;ms^{β1}\), \(v_o\)β=\(0\) \(ms^{β1}\) and \(v_s\)β = 4\(\sqrt{2}\) msβ1, let's calculate the observed frequency at \(π\) when only \(π_2\)emits sound
at \(π\).\(f'=656\times\frac{324+0}{324+4\sqrt{2}}\)
\(f'=656\times\frac{324}{324+4\sqrt{2}}\)
\(f' = 648.00\)
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Concepts Used:
Oscillations
Oscillation is a process of repeating variations of any quantity or measure from its equilibrium value in time . Another definition of oscillation is a periodic variation of a matter between two values or about its central value.
The term vibration is used to describe the mechanical oscillations of an object. However, oscillations also occur in dynamic systems or more accurately in every field of science. Even our heartbeats also creates oscillationsβ. Meanwhile, objects that move to and fro from its equilibrium position are known as oscillators.
Read More: Simple Harmonic Motion
Oscillation- Examples
The tides in the sea and the movement of a simple pendulum of the clock are some of the most common examples of oscillations. Some of examples of oscillations are vibrations caused by the guitar strings or the other instruments having strings are also and etc. The movements caused by oscillations are known as oscillating movements. For example, oscillating movements in a sine wave or a spring when it moves up and down.
The maximum distance covered while taking oscillations is known as the amplitude. The time taken to complete one cycle is known as the time period of the oscillation. The number of oscillating cycles completed in one second is referred to as the frequency which is the reciprocal of the time period.