
A train is moving at a speed of 220 m/s towards a stationary object, emits a sound of frequency 1000Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (speed off the sound in air is 330 m/s)
A. 4000Hz
B. 5000Hz
C. 3000Hz
D. 3500Hz
Answer
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Hint: Doppler effect is known as, when the source and observer are moving relative to each other, the frequency observed by the observer ‘ ${f_a}$’ is different from the actual frequency produced by the source ‘ ${f_ \circ }$’. The general formula of doppler effect is given as, $f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$.
Formula Used: $f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$
Complete answer:
Frequency is defined as the estimation of the number of times that a repeated event occurs per unit of time. The frequency of wave-like patterns such as the electromagnetic waves i.e. radio waves or light waves, sound waves, electrical signals, or other waves, expresses the number of cycles of the repetitive waveform per second.
According to the concept of doppler effect, when an observer moves towards a stationary source of sound then the apparent frequency heard by the observer increases. So, the apparent frequency heard in this situation is given as,
$f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$
It is given that the,
Speed of the train = 220 m/s
Speed of the sound in the air = 330 m/s
By applying the doppler effect frequency of the echo which is detected by the driver of the train is given as,
$\eqalign{
& \Rightarrow f' = \left( {\dfrac{{330 + 220}}{{330 - 220}}} \right)1000 \cr
& \Rightarrow f' = 1000\left( {\dfrac{{550}}{{110}}} \right) \cr
& \therefore f' = 5000Hz \cr} $
Thus, the frequency of the echo as detected by the driver of the train is 5000Hz.
Hence, option (B) is the correct answer.
Note:
Students tend to get mistaken that doppler effect is only applicable for sound waves. But, in reality it works with all types of wave forms including light. Some of the applications of doppler effect are:
1) It is used in sirens, radars, and in satellite communications.
2) It is also used for medical purposes such as medical imaging, blood flow measurement.
Formula Used: $f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$
Complete answer:
Frequency is defined as the estimation of the number of times that a repeated event occurs per unit of time. The frequency of wave-like patterns such as the electromagnetic waves i.e. radio waves or light waves, sound waves, electrical signals, or other waves, expresses the number of cycles of the repetitive waveform per second.
According to the concept of doppler effect, when an observer moves towards a stationary source of sound then the apparent frequency heard by the observer increases. So, the apparent frequency heard in this situation is given as,
$f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$
It is given that the,
Speed of the train = 220 m/s
Speed of the sound in the air = 330 m/s
By applying the doppler effect frequency of the echo which is detected by the driver of the train is given as,
$\eqalign{
& \Rightarrow f' = \left( {\dfrac{{330 + 220}}{{330 - 220}}} \right)1000 \cr
& \Rightarrow f' = 1000\left( {\dfrac{{550}}{{110}}} \right) \cr
& \therefore f' = 5000Hz \cr} $
Thus, the frequency of the echo as detected by the driver of the train is 5000Hz.
Hence, option (B) is the correct answer.
Note:
Students tend to get mistaken that doppler effect is only applicable for sound waves. But, in reality it works with all types of wave forms including light. Some of the applications of doppler effect are:
1) It is used in sirens, radars, and in satellite communications.
2) It is also used for medical purposes such as medical imaging, blood flow measurement.
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