The document discusses the relationships among power, intensity, and displacement amplitude in sound waves. It explains how power affects amplitude and provides equations to determine these relationships. Additionally, it illustrates concepts through problem-solving scenarios related to sound wave behavior and properties.

1. Sound waves:
Relating displacement
amplitude, power
and intensity
Alyssa Hui
Phys 101 201
LO4

2. • Power (P) is the rate at which the energy is
transferred by a wave, with units of J/s
• Intensity is the power delivered per unit area,
giving units of W/m2
• Intensity is also determined by the density of
the medium, wave speed, angular frequency
and the displacement amplitude (sm)
Concepts

3. • When we double the power produced
by a source, by what factor does the
displacement amplitude of the resultant
wave change?
Problem 1
cr: desmos.com

4. •I = P/A
•I = ½ ρvω2sm
2
Relevant Equations

5. • Can you find the relationship
between sm and P?
Hint

6. • Setting the two equations for intensity equal
to each other, we can solve for sm,
½ ρvω2sm
2 = P/A
sm = (2𝑃/(𝐴𝜌𝑣𝜔2))
• If P is doubled,
sm = (2 ∗ (2𝑃))
the displacement amplitude of the
resultant wave increases by approx. 1.4.
Solution

7. • The equation
sm = (2𝑃/(𝐴𝜌𝑣𝜔2))
can be used to determine relationships
between other properties in a sound
wave, such as in the following problem.
Note

8. • A person is standing beside a speaker
as it plays a 10,000 Hz tone. The sound
waves travel away from the speaker
uniformly in all directions. If the
distance from the speaker doubles,
then the amplitude of the waves that
the person observes:
Problem 2

9. • (From previous problem)
sm = (2𝑃/(𝐴𝜌𝑣𝜔2))
Relevant Equation

10. •What area does power
radiating isotropically cover?
Hint

11. • Radiating isotropically means waves will travel
uniformly in all directions; therefore power is
radiated over A = 4πr2,
sm = (2𝑃/(4πr2 ∗ 𝜌𝑣𝜔2))
• With distance doubled, radius r is doubled,
sm = (1/ 2𝑟 2)
and the amplitude of the waves is halved.
Solution