Southern New Hampshire University
PHY-150 - Physics I with Lab
PHY-150 Module Eight Homework | Questions &
Complete Solutions
Wave Mechanics, Sound, Fluid Dynamics, Thermodynamics | SNHU | Title 63 | 2026 Update
Assessment Module Eight Homework - Questions & Complete Solutions (Title 63)
Questions 25 Multiple Choice (A-D) with Step-by-Step Solutions
Date July 8, 2026
Academic Year 2026-2027
Cognitive Level 30% Recall | 50% Application | 20% Analysis
Standards SNHU PHY-150 Module Eight Competencies, 2026 Physics Standards
CONFIDENTIALITY NOTICE: This document is prepared exclusively for Southern New Hampshire University academic use.
Unauthorized distribution, reproduction, or sale is strictly prohibited.
, SNHU PHY-150 | Module Eight Homework | Title 63 | 2026
Section 1: Wave Mechanics, Oscillations, & Sound
Q1. A 0.50 kg mass is attached to a spring with spring constant k = 200 N/m and displaced from equilibrium. Calculate
the period T and frequency f of the resulting simple harmonic motion.
A. T = 0.628 s, f = 1.59 Hz
B. T = 0.050 s, f = 20.0 Hz
C. T = 0.314 s, f = 3.18 Hz [CORRECT]
D. T = 1.00 s, f = 1.00 Hz
Correct Answer: C
Rationale: T = 2π√(m/k) = 2π√(0.50/200) = 2π√(0.00250) = 2π(0.0500) = 0.314 s. f = 1/0.314 = 3.18 Hz. A inverts the
ratio under the square root (k/m instead of m/k). B uses an invalid algebraic form. D uses an unsourced formula.
Q2. A steel guitar string under 50.0 N of tension has a linear mass density of 2.00 × 10-3 kg/m. What is the wave speed
on the string?
A. 158 m/s [CORRECT]
B. 25.0 m/s
C. 500 m/s
D. 100 m/s
Correct Answer: A
Rationale: v = √(T/µ) = √(50.0/0.00200) = √(2.50 × 104) = 158 m/s. B forgot the square root (v = T/µ = 25,000 m/s, then
misinterpreted). C and D use incorrect formulas.
Q3. An ambulance siren emitting at 440 Hz moves toward a stationary observer at 30.0 m/s. If the speed of sound is 343
m/s, what frequency does the observer hear?
A. 404 Hz
B. 440 Hz
C. 472 Hz
D. 482 Hz [CORRECT]
Correct Answer: D
Rationale: For source moving toward stationary observer: f' = f·v/(v - vs) = 440(343)/(343 - 30) = 440(1.096) = 482 Hz. A
uses the formula for source moving away. B ignores Doppler effect. C uses wrong sign (v + vs instead of v - vs).
Q4. A string of length 0.65 m is fixed at both ends. If the wave speed on the string is 158 m/s, what is the fundamental
frequency (first harmonic)?
A. 243 Hz
B. 121 Hz [CORRECT]
C. 60.6 Hz
D. 486 Hz
Correct Answer: B
Rationale: For a string fixed at both ends, fn = n·v/(2L). f1 = v/(2L) = 158/(2 × 0.65) = 158/1.30 = 121.5 Hz. A forgets the
factor of 2 (uses v/L). C uses the closed-pipe formula. D inverts the factor of 2.
Q5. A sound wave has an intensity of 1.0 × 10-4 W/m². What is the sound intensity level in decibels? (I0 = 1.0 × 10-12
W/m²)
A. 40.0 dB
B. 4.0 dB
C. 80.0 dB [CORRECT]
D. 100 dB
Correct Answer: C
Rationale: β = 10·log(I/I0) = 10·log(10-4/10-12) = 10·log(108) = 10 × 8 = 80 dB. A forgets to divide by I0. B omits the
factor of 10. D miscalculates the ratio.
Q6. What is the speed of sound in air at a temperature of 25.0 °C?
A. 346 m/s [CORRECT]
B. 331 m/s
C. 509 m/s
Page 2
PHY-150 - Physics I with Lab
PHY-150 Module Eight Homework | Questions &
Complete Solutions
Wave Mechanics, Sound, Fluid Dynamics, Thermodynamics | SNHU | Title 63 | 2026 Update
Assessment Module Eight Homework - Questions & Complete Solutions (Title 63)
Questions 25 Multiple Choice (A-D) with Step-by-Step Solutions
Date July 8, 2026
Academic Year 2026-2027
Cognitive Level 30% Recall | 50% Application | 20% Analysis
Standards SNHU PHY-150 Module Eight Competencies, 2026 Physics Standards
CONFIDENTIALITY NOTICE: This document is prepared exclusively for Southern New Hampshire University academic use.
Unauthorized distribution, reproduction, or sale is strictly prohibited.
, SNHU PHY-150 | Module Eight Homework | Title 63 | 2026
Section 1: Wave Mechanics, Oscillations, & Sound
Q1. A 0.50 kg mass is attached to a spring with spring constant k = 200 N/m and displaced from equilibrium. Calculate
the period T and frequency f of the resulting simple harmonic motion.
A. T = 0.628 s, f = 1.59 Hz
B. T = 0.050 s, f = 20.0 Hz
C. T = 0.314 s, f = 3.18 Hz [CORRECT]
D. T = 1.00 s, f = 1.00 Hz
Correct Answer: C
Rationale: T = 2π√(m/k) = 2π√(0.50/200) = 2π√(0.00250) = 2π(0.0500) = 0.314 s. f = 1/0.314 = 3.18 Hz. A inverts the
ratio under the square root (k/m instead of m/k). B uses an invalid algebraic form. D uses an unsourced formula.
Q2. A steel guitar string under 50.0 N of tension has a linear mass density of 2.00 × 10-3 kg/m. What is the wave speed
on the string?
A. 158 m/s [CORRECT]
B. 25.0 m/s
C. 500 m/s
D. 100 m/s
Correct Answer: A
Rationale: v = √(T/µ) = √(50.0/0.00200) = √(2.50 × 104) = 158 m/s. B forgot the square root (v = T/µ = 25,000 m/s, then
misinterpreted). C and D use incorrect formulas.
Q3. An ambulance siren emitting at 440 Hz moves toward a stationary observer at 30.0 m/s. If the speed of sound is 343
m/s, what frequency does the observer hear?
A. 404 Hz
B. 440 Hz
C. 472 Hz
D. 482 Hz [CORRECT]
Correct Answer: D
Rationale: For source moving toward stationary observer: f' = f·v/(v - vs) = 440(343)/(343 - 30) = 440(1.096) = 482 Hz. A
uses the formula for source moving away. B ignores Doppler effect. C uses wrong sign (v + vs instead of v - vs).
Q4. A string of length 0.65 m is fixed at both ends. If the wave speed on the string is 158 m/s, what is the fundamental
frequency (first harmonic)?
A. 243 Hz
B. 121 Hz [CORRECT]
C. 60.6 Hz
D. 486 Hz
Correct Answer: B
Rationale: For a string fixed at both ends, fn = n·v/(2L). f1 = v/(2L) = 158/(2 × 0.65) = 158/1.30 = 121.5 Hz. A forgets the
factor of 2 (uses v/L). C uses the closed-pipe formula. D inverts the factor of 2.
Q5. A sound wave has an intensity of 1.0 × 10-4 W/m². What is the sound intensity level in decibels? (I0 = 1.0 × 10-12
W/m²)
A. 40.0 dB
B. 4.0 dB
C. 80.0 dB [CORRECT]
D. 100 dB
Correct Answer: C
Rationale: β = 10·log(I/I0) = 10·log(10-4/10-12) = 10·log(108) = 10 × 8 = 80 dB. A forgets to divide by I0. B omits the
factor of 10. D miscalculates the ratio.
Q6. What is the speed of sound in air at a temperature of 25.0 °C?
A. 346 m/s [CORRECT]
B. 331 m/s
C. 509 m/s
Page 2
