AP Physics 2 Exam Question

A loudspeaker is emitting sound waves with a frequency of 800 Hz. The sound waves travel through air with a speed of 343 m/s.

a) Calculate the wavelength of these sound waves. Show all your work. b) If each sound wave has a peak amplitude of 2.5 Pa, determine the sound intensity level at a distance of 5.0 meters from the loudspeaker.

---

Part a)

To calculate the wavelength of the sound wave, we can use the formula:

$\lambda = \frac{v}{f}$

where: $\lambda$ = wavelength of the sound wave $v$ = speed of sound in air (343 m/s) $f$ = frequency of the sound wave (800 Hz)

Substituting the given values into the equation:

$\lambda = \frac{343 \, \text{m/s}}{800 \, \text{Hz}}$

Calculating this expression using appropriate units:

$\lambda = \frac{343 \, \text{m/s}}{800 \, \text{s}^{-1}}$

$\lambda = 0.428 \, \text{m}$

Therefore, the wavelength of the sound waves is 0.428 m.

Part b)

To determine the sound intensity level at a distance of 5.0 meters from the loudspeaker, we can use the formula:

$L = 10 \log\left(\frac{I}{I_0}\right)$

where: $L$ = sound intensity level in decibels (dB) $I$ = intensity of the sound wave at the given distance (unknown) $I_0$ = reference intensity of sound wave (typically set to $10^{-12} \, \text{W/m}^2$)

To find the intensity $I$ at a distance of 5.0 meters from the loudspeaker, we first need to calculate the initial intensity $I_1$ at a distance of 0.428 m (wavelength) from the loudspeaker. We can use the formula:

$\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2$

where: $I_1$ = initial intensity at a distance of 0.428 m $I_2$ = final intensity at a distance of 5.0 m $r_1$ = initial distance from the loudspeaker (0.428 m) $r_2$ = final distance from the loudspeaker (5.0 m)

Substituting the given values into the equation:

$\frac{I_1}{I_2} = \left(\frac{5.0 \, \text{m}}{0.428 \, \text{m}}\right)^2$

Calculating this expression:

$\frac{I_1}{I_2} = \left(\frac{11.682}{0.428}\right)^2$

$\frac{I_1}{I_2} = 308.443$

Now, we can use the relationship between intensity and amplitude:

$I_1 = \frac{1}{2}\rho v A^2 f$

where: $\rho$ = air density (typically assumed to be 1.225 kg/m^3) $v$ = speed of sound in air (343 m/s) $A$ = amplitude of the sound wave (2.5 Pa) $f$ = frequency of the sound wave (800 Hz)

Substituting the given values into the equation:

$I_1 = \frac{1}{2} \times 1.225 \, \text{kg/m}^3 \times 343 \, \text{m/s} \times (2.5 \, \text{Pa})^2 \times 800 \, \text{Hz}$

Calculating this expression:

$I_1 = 5273000 \, \text{W/m}^2$

Now, we can find the final intensity $I_2$ at a distance of 5.0 meters from the loudspeaker:

$I_2 = \frac{I_1}{308.443}$

Calculating this expression:

$I_2 = \frac{5273000 \, \text{W/m}^2}{308.443}$

$I_2 = 17100 \, \text{W/m}^2$

Finally, we can calculate the sound intensity level $L$ at a distance of 5.0 meters using the formula:

$L = 10 \log\left(\frac{I_2}{I_0}\right)$

Substituting the given values into the equation:

$L = 10 \log\left(\frac{17100 \, \text{W/m}^2}{10^{-12} \, \text{W/m}^2}\right)$

Calculating this expression:

$L = 10 \log(1.71 \times 10^{16})$

$L = 163.3 \, \text{dB}$

Therefore, the sound intensity level at a distance of 5.0 meters from the loudspeaker is 163.3 dB.

---

Note: Remember to provide units throughout your calculations and round your final answers to the appropriate number of significant figures.