Question

The distance between a loudspeaker and the left ear of a listener is 2.70 m. (a) Calculate the time required for sound to travel this distance if the air temperature is 20 oc. (b) Assuming that the sound frequency is 523 Hz, how many wavelengths of sound are contained in this distance

Answer 1

Answer:

Explanation:

Distance covered by sound = 2.7 m

speed of sound at 20⁰C = 343 m /s

time take by sound to cover the distance = distance / speed

= 2.7 / 343

= 7.87 ms . ( millisecond )

b )

Wavelength of sound = speed / frequency

= 343 / 523 m

= .6558 m

= 65.58 cm

Distance = 2.70 m = 270 cm

No of wavelengths contained in the distance

= 270 / 65.58

= 4.11 or 4 wavelengths ( by rounding off to digit )

Answer 2

The time taken by sound and the wavelengths of sound is required.

Time taken is $0.00787\ \text{s}$

The number of wavelenghts is 4.

s = Distance = 2.7 m

v = Speed of sound at $20^{\circ}\text{C}$ = 343 m/s

Time is given by

$t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{2.7}{343}\\\Rightarrow t=0.00787\ \text{s}$

f = Frequency = 523 Hz

Wavelength is given by

$\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{523}\\\Rightarrow \lambda=0.66\ \text{m}$

The wavelength is 0.66 m.

n = Number of wavelengths

$n\lambda=s\\\Rightarrow n=\dfrac{s}{\lambda}\\\Rightarrow n=\dfrac{2.7}{0.66}\\\Rightarrow n\approx 4$

The number of wavelenghts is 4.

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