In the picture below, the total mass of the speaker and the block it is attached

Question

In the picture below, the total mass of the speaker and the block it is attached to is m = 0.50 kg. The spring has a force constant k = 3200 N/m. The speaker emits a tone at a frequency = 8.0 kHz. When the spring is at its equilibrium length the distance between the speaker and receiver is d= 1.9 m. Also, when at equilibrium, the intensity level of the tone at the receiver is beta = 40 dB. The block is pulled a distance A = 0.44 m to the right and released. The block and attached speaker begin moving back and forth in simple harmonic motion. Take the speed of sound to be v = 340 m/s.

Explanation / Answer

Omega, w = sqrt(k/m) = sqrt(3200/0.5) = 80 rad/s2

Vmax = amplitude *w = 80*0.44 = 35.2 m/s

Now, the receiver will get the aparent frequency, the range of which will be given by.

f' = f*V/(V - Vs) or f' = f*V/(V + Vs)

f' max = 8.923 kHz ; f' min = 7.249 kHz

b) Decible in terms of density is given by,

decible = 10 log (d/dref)^2 ; d = density and dref = reference density.

d min = 1.9-0.44 = 1.46 m; d max = 1.9+0.44 = 2.34 m

I proportional to 1/d^2

I * d^2 = I min * dmax^2 = I max * dmin^2 = k

I = kd*d = k/3.61

I min = k/5.4756

I max = k/2.1316

At equi, dB = 40.

dB = 10 log(I/Io) ; Io = 10^-14 ; So, I = 10^-10

So, k = I*3.61 = 3.61*10^-10

I max = 19.767*10^-10

I min = 7.695*10^-10

In terms of decible, dB max = 52.96 ; db min = 48.86

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