A source of the sound of frequency $f_{1}$ is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency f(Hz) is plotted against time t(sec). The speed of sound in air is 300 m/s. Find $f_{1} (g = 10 m / s^{2})$ .

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Solution

The correct option is B
$f = (\frac{V + V_{0}}{V}) f_{1} = f_{1} + f_{1} \frac{V_{0}}{V} V_{0} = g t S o, f = f_{1} + (\frac{f_{1} g}{V}) t S l o p e o f g r a p h = \frac{f_{1} g}{V} \frac{(2) (10^{- 3}) - f_{1}}{30} = \frac{(f_{1}) (10)}{300} o r f_{1} = 10^{3} H z$

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Similar questions

Q. A source of the sound of frequency

f_{1}

is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency f(Hz) is plotted against time t(sec). The speed of sound in air is 300 m/s. Find

f_{1} (g = 10 m / s^{2})

Q. A detector is released from rest over a source of sound of frequency

f_{0} = 10^{3} H z

. The frequency observed by the detector at time t is plotted in the graph. The speed of sound in air is:

(g = 10 m / s^{2})

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a source of sound frequency 600 hz is placed inside water .the speed of sound in water is 1500m/s and in air is 300m/s.the frequency if sound recorded by an observer ,who is standing in air is.

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A source of the sound of frequency f1 is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency f(Hz) is plotted against time t(sec). The speed of sound in air is 300 m/s. Find f1(g=10m/s2).

The correct option is B f=(V+V0V)f1=f1+f1V0VV0=gtSo,f=f1+(f1gV)tSlopeofgraph=f1gV(2)(10−3)−f130=(f1)(10)300orf1=103Hz

A source of the sound of frequency $f_{1}$ is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency f(Hz) is plotted against time t(sec). The speed of sound in air is 300 m/s. Find $f_{1} (g = 10 m / s^{2})$ .

The correct option is B
$f = (\frac{V + V_{0}}{V}) f_{1} = f_{1} + f_{1} \frac{V_{0}}{V} V_{0} = g t S o, f = f_{1} + (\frac{f_{1} g}{V}) t S l o p e o f g r a p h = \frac{f_{1} g}{V} \frac{(2) (10^{- 3}) - f_{1}}{30} = \frac{(f_{1}) (10)}{300} o r f_{1} = 10^{3} H z$