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3 years ago

5

Sound wave W has amplitude ‘A' and frequency ‘f'. Sound wave X is louder and lower in pitch than sound wave W. What can be said

about the amplitude and frequency of sound wave X?

1 answer:

ra1l [238]3 years ago

7 0

Answer:

Sound wave X amplitude is greater than 'A' and its frequency is lesser than

'f'

Explanation:

The pitch of a sound is dictated by the frequency of the sound wave, while the loudness is dictated by the amplitude.

A high pitch sound corresponds to a high frequency and a low pitch sound corresponds to a low frequency.

The larger the amplitude of the waves, the louder the sound and vice-versa.

From the question,

Sound wave W has amplitude ‘A' and frequency 'f' and

Sound wave X is louder and lower in pitch than sound wave W.

Since sound wave X is louder, this means its amplitude is greater than 'A'.

Also, since sound wave X is lower in pitch, this means its frequency is lesser than 'f'.

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we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

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This value can be used to determine the time for the pellet through the barrel:

$t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s$

Finally, we can use the above to calculate the force:

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3 years ago

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

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Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

Answer:

a

$\Delta f = 81.93 \ Hz$

b

$\lambda_1 = 0.867 \ m$

Explanation:

From the question we are told that

The speed of the train is $v_t = 39.6 m/s$

The frequency of the train horn is $f_t = 350 \ Hz$

Generally the speed of sound has a constant values of $v_s = 343 m/s$

Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_1 = f * \frac{v_s}{v_s - v_t}$

substituting values

$f_1 = 350 * \frac{343 }{343-39.6}$

$f_1 = 395.7 \ Hz$

Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_2 = f * \frac{v_s}{v_s +v_t}$

substituting values

$f_2 = 350 * \frac{343}{343 + 39.6}$

$f_2 = 313.77 \ Hz$

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

$\Delta f = f_1 - f_2$

$\Delta f = 395.7 - 313.77$

$\Delta f = 81.93 \ Hz$

Generally the wavelength detected by the person as the train approaches is mathematically represented as

$\lambda_1 = \frac{v}{f_1 }$

$\lambda_1 = \frac{343}{395.7 }$

$\lambda_1 = 0.867 \ m$

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