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Sound wave properties and calculations.

Created by @nathanedwards
 at November 1st 2023, 8:09:31 am.
AP_Physics_2-Questions
#sound
#intensity
#wave

Question:

A sound wave of frequency 440 Hz and wavelength 0.775 m is traveling through a medium with a speed of 330 m/s. The wave is produced by a tuning fork.

a) Calculate the period of the sound wave.

b) Determine the amplitude of the wave if the maximum displacement of a particle in the medium is 0.25 cm.

c) The wave encounters a medium with a density of 1.2 kg/m^3. Calculate the intensity of the sound wave.

d) The sound wave now propagates through a different medium with a speed of 400 m/s and a density of 2 kg/m^3. Determine the new wavelength of the sound wave.

e) Given that the sound wave has an intensity level of 80 dB at a certain point, calculate the corresponding sound intensity in watts per square meter.

Answer:

a) The formula for the period (T) of a wave is given by the reciprocal of the frequency (f):

T=1f T = \frac{1}{f}

Given that the frequency is 440 Hz, we can substitute the value into the formula:

T=1440=0.00227 s T = \frac{1}{440} = 0.00227 \, \text{s}

Therefore, the period of the sound wave is 0.00227 seconds.

b) The displacement amplitude (A) of a wave is related to the maximum displacement (Dmax) of the particles in the medium. Since the maximum displacement is given in centimeters, we need to convert it to meters:

Dmax=0.25 cm×(1 m100 cm)=0.0025 m D_{\text{{max}}} = 0.25 \, \text{{cm}} \times \left(\frac{1 \, \text{{m}}}{100 \, \text{{cm}}}\right) = 0.0025 \, \text{{m}}

Therefore, the amplitude of the wave is 0.0025 meters.

c) The intensity (I) of a wave is given by the formula:

I=PA I = \frac{{P}}{{A}}

Where P is the power carried by the wave and A is the area through which the wave travels. The power can be calculated using the formula:

P=ρ⋅v⋅f⋅A2⋅λ2 P = \frac{{\rho \cdot v \cdot f \cdot A^2 \cdot \lambda}}{{2}}

where rho (ρ) is the density of the medium, v is the velocity of the wave, f is the frequency of the wave, A is the amplitude of the wave, and lambda (λ) is the wavelength of the wave.

Substituting the given values into the formula:

I=(1.2 kg/m3)⋅(330 m/s)⋅(440 Hz)⋅(0.0025 m)2⋅0.775 m2 I = \frac{{(1.2 \, \text{{kg/m}}^3) \cdot (330 \, \text{{m/s}}) \cdot (440 \, \text{{Hz}}) \cdot (0.0025 \, \text{{m}})^2 \cdot 0.775 \, \text{{m}}}}{{2}}
I≈31.245 W/m2 I \approx 31.245 \, \text{{W/m}}^2

Therefore, the intensity of the sound wave is approximately 31.245 Watts per square meter.

d) The relationship between the speed of the wave (v), frequency (f), and wavelength (λ) is given by the formula:

v=f⋅λ v = f \cdot \lambda

Rearranging the formula to solve for the wavelength (λ):

λ=vf=400 m/s440 Hz≈0.909 m \lambda = \frac{{v}}{{f}} = \frac{{400 \, \text{{m/s}}}}{{440 \, \text{{Hz}}}} \approx 0.909 \, \text{{m}}

Therefore, the new wavelength of the sound wave is approximately 0.909 meters.

e) The formula for calculating the sound intensity level (L) given the intensity (I) is:

L=10⋅log⁡10(II0) L = 10 \cdot \log_{10}\left(\frac{{I}}{{I_0}}\right)

Where I_0 is the reference intensity, which is commonly taken as 10^(-12) Watts per square meter.

Rearranging the formula to solve for the sound intensity (I):

I=I0⋅10(L10) I = I_0 \cdot 10^{\left(\frac{{L}}{{10}}\right)}

Given that the sound intensity level is 80 dB, we can substitute the values into the formula:

I=(10−12 W/m2)⋅10(8010)=10−2 W/m2 I = (10^{-12} \, \text{{W/m}}^2) \cdot 10^{\left(\frac{{80}}{{10}}\right)} = 10^{-2} \, \text{{W/m}}^2

Therefore, the sound intensity is 10^(-2) Watts per square meter.

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