Purpose: Test understanding of wave properties, sound, resonance, harmonics, and the Doppler effect.
Formulae: \( v=f\lambda \) | \( T=1/f \) | \( v_\text{sound}=331+0.6T_C \) m/s | String: \( f_n=nv/(2L) \) | Open pipe: \( f_n=nv/(2L) \) | Closed pipe: \( f_n=nv/(4L) \), n odd | Doppler: \( f' = f \cdot v/(v-v_s) \) (approaching)
Score: 0 / 10
Q1 — Wave equation
A wave on a string has frequency \(50\) Hz and wavelength \(0.40\) m. What is its speed?
Answer: m/s
Solution:
\( v = f\lambda = 50(0.40) = 20 \) m/s
Q2 — Speed of sound in air
Calculate the speed of sound at \(20°\)C.
Answer: m/s
Solution:
\( v = 331 + 0.6(20) = 331 + 12 = 343 \) m/s
Q3 — Period and frequency
A pendulum completes 30 oscillations in 60 seconds. What is its frequency?
Answer: Hz
Solution:
\( f = 30/60 = 0.5 \) Hz (or \( T = 2 \) s, \( f = 1/T = 0.5 \) Hz)
Q4 — Echo timing
A sonar pulse returns from a fish 0.40 s after being emitted. The water temperature is such that sound speed is \(1500\) m/s. How far away is the fish?
Answer: m
Solution:
Round-trip distance: \( d_\text{total} = vt = 1500(0.40) = 600 \) m. One-way: \( 300 \) m.
Q5 — Wavelength of sound
A guitar string vibrates at \(440\) Hz (concert A) in air at \(20°\)C. What is the wavelength of the sound it produces in air?
Answer: m
Solution:
\( v = 343 \) m/s; \( \lambda = v/f = 343/440 = 0.78 \) m
Q6 — Standing wave on string
A string of length \(0.50\) m is fixed at both ends and vibrates in its fundamental mode (n=1). The wave speed on the string is \(200\) m/s. What is the fundamental frequency?
Answer: Hz
Solution:
\( f_1 = v/(2L) = 200/(2\times 0.50) = 200 \) Hz
Q7 — Closed pipe (resonance)
A pipe closed at one end has length \(0.85\) m. What is the fundamental frequency of standing wave produced inside (\(v=343\) m/s)?
Answer: Hz
Solution:
For a closed pipe (n odd): \( f_1 = v/(4L) = 343/(4\times 0.85) = 343/3.4 = 101 \) Hz
Q8 — Doppler effect (approaching source)
A train sounds its horn at \(500\) Hz while approaching a stationary observer at \(34\) m/s. Speed of sound is \(343\) m/s. What frequency does the observer hear?
Answer: Hz
Solution:
\( f' = f \cdot \frac{v}{v-v_s} = 500 \cdot \frac{343}{343-34} = 500 \cdot \frac{343}{309} = 555 \) Hz
Q9 — Wave concept
When a sound wave travels from air into water, which property of the wave changes most?
Solution:
Frequency is set by the source and does not change as the wave moves between media. Speed depends on the medium (faster in water). Since \( v=f\lambda \), wavelength changes proportionally to \( v \).
Q10 — Beat frequency
Two tuning forks at \(440\) Hz and \(443\) Hz are sounded together. What is the beat frequency the listener hears?