Question

A bolt of lightening strikes a building in the city. You are on top of a hill 15 km away.

1. How much longer does it take the sound of the thunder created by the lightning to reach your ears than the light from the lightning to reach your eyes if the sound waves travel at a speed of 363 m/s and light travels at a speed of c = 3.00 x 108 m/s?

2. If the dominant frequency of the sound wave is 63 Hz, what is the wavelength ofthe sound wave?

3.If the sound level of the thunder on the top of the hill is 42.0 dB, what is the intensity of the sound wave on the hill?

4.Assuming the lightning acts like a point source for sound waves, what would be the intensity of the sound wave from the same lightning bolt at a location which is only 0.400 km away from the city?

5.What would be the sound level (in dB) of the thunder at a location which is only 0.400 km away?

6.What would be the sound level (in dB) at a location which is only 0.500 km away if two identical lightning bolts struck at the same time? Would this sound level be above the pain threshold?

Answer 1

1) time diffrence reached by sound and light = d/v_sound - d/c

= 15*10^3/363 - 15*10^3/(3*10^8)

= 41.32226 s

2) use, v = lamda*f

==> lamda = v/f

= 363/63

= 5.76 m

3) we know, sound intensity level

beta = 10*log(I/Io)

42 = 10*log(I/Io)

4.2 = log(I/Io)

10^4.2 = I/Io

I = Io*10^4.2

= 10^-12*10^4.2

= 1.58*10^-8 W/m^2

4) I2/I1 = (r1/r2)^2

I2 = I1*(r1/r2)^2

= 1.58*10^-8*(15/0.4)^2

= 2.22*10^-5 W/m^2

5) sound intensity level, beta = 10*log(I2/Io)

= 10*log(2.22*10^-5/10^-12)

= 73.5 dB

6) intensity of sound at 0.5 km,

I = 2.22*10^-5*(0.4/0.5)^2

= 1.4208*10^-5 W/m^2

sound intensity level, beta = 10*log(2*I/Io)

= 10*log(2*1.4208*10^-5/10^-12)

= 74.5 dB