Question

a stationary source produces a sound wave at a frequency of 100 hz. the wave travels at 925 feet per second. a car is moving toward the source at a speed of 25 feet per second. what is the wavelength of the stationary sound source and the wavelength that a person in the car perceives? (1 point) wavelength of the stationary source: 9.25 perceived wavelength: 9.00 ft wavelength of the stationary source: 11.25 perceived wavelength: 9.25 ft wavelength of the stationary source: 10.25 perceived wavelength: 9.50 ft wavelength of the stationary source: 11.25 perceived wavelength: 13.25 ft

Explanation:

Step1: Calculate stationary wavelength

Use the formula $\lambda = \frac{v}{f}$, where $v = 925$ ft/s and $f = 100$ Hz.
$\lambda = \frac{925}{100} = 9.25$ ft.

Step2: Calculate perceived wavelength (Doppler effect)

When the observer (car) moves toward the source, the perceived frequency $f'$ is $f' = f \cdot \frac{v + v_o}{v}$, where $v_o = 25$ ft/s.
First, find $f'$: $f' = 100 \cdot \frac{925 + 25}{925} = 100 \cdot \frac{950}{925} \approx 102.70$ Hz (approx, but use $\lambda' = \frac{v}{f'}$).
Alternatively, $\lambda' = \frac{v - v_o}{f}$ (since observer moving toward source, relative speed is $v - v_o$ for wavelength perception? Wait, correct formula: for observer moving toward source, the perceived wavelength is $\lambda' = \frac{v - v_o}{f}$? Wait, no: wavelength is distance between crests. If observer moves toward source, the time between crests is less, so perceived frequency increases, wavelength decreases.

Wait, correct approach: $\lambda = \frac{v}{f} = 9.25$ ft (stationary). For moving observer toward source, the relative speed of waves with respect to observer is $v + v_o$? No, wave speed is $v$ (medium speed), observer speed is $v_o$ toward source. So the rate at which observer encounters crests is $f' = \frac{v + v_o}{\lambda}$, but $\lambda = \frac{v}{f}$, so $f' = f \cdot \frac{v + v_o}{v}$. Then perceived wavelength $\lambda' = \frac{v}{f'} = \frac{v}{f \cdot \frac{v + v_o}{v}} = \frac{v^2}{f(v + v_o)}$? Wait, no, that's wrong. Wait, original wavelength is $\lambda = v/f$. When observer moves toward source, the distance between crests as perceived is $\lambda' = \lambda - v_o \cdot T$, where $T = 1/f$ is period. So $\lambda' = \frac{v}{f} - \frac{v_o}{f} = \frac{v - v_o}{f}$. Let's check: $v = 925$, $v_o = 25$, $f = 100$. So $\lambda' = \frac{925 - 25}{100} = \frac{900}{100} = 9.00$ ft. Ah, that's simpler! Because in time $T = 1/f$, the wave moves $vT = \lambda$ ft, and the observer moves $v_o T$ ft toward the source, so the distance between crests (perceived wavelength) is $\lambda - v_o T = \lambda - \frac{v_o}{f} = \frac{v}{f} - \frac{v_o}{f} = \frac{v - v_o}{f}$.

So $\lambda' = \frac{925 - 25}{100} = \frac{900}{100} = 9.00$ ft.

So stationary wavelength is 9.25 ft, perceived is 9.00 ft.

Answer:

wavelength of the stationary source: 9.25 ft, perceived wavelength: 9.00 ft (the first option)