Power is the rate at which energy is delivered by a wave. Intensity is the power per unit area of the wave. Formulas show that intensity is equal to power divided by area. Sample problems demonstrate calculating power from intensity and area, as well as calculating amplitude from known values like period, pressure amplitude, and bulk modulus. A clicker question example finds the length of a square plate needed to receive a given intensity from a speaker of known power. Another clicker question example calculates the intensity on a basketball from a sound wave of known amplitude, frequency, and the basketball's radius.

1. Power and Intensity
A PHYSICS 101 201 LEARNING OBJECT 4 BY NICO UTAMA ALIMIN

2. What is Power?
Power is the rate at which the wave delivers energy
◦ Energy per unit time
◦ In Joules/second or Watts
What is Intensity?
• Intensity of a wave is power per unit area
• Average force exerted per unit time per unit area
• In
𝐽𝑜𝑢𝑙𝑒𝑠
𝑆𝑒𝑐𝑜𝑛𝑑 𝑀𝑒𝑡𝑒𝑟𝑠2 or
𝑊𝑎𝑡𝑡𝑠
𝑀𝑒𝑡𝑒𝑟𝑠2

3. Formulas?
I =
𝑃
𝐴
I = Intensity
P = Power
A = Area
 Average Power:
 Pavg =
1
2
𝜇𝑣𝜔2A2
 𝜇 = Linear Mass Density
 v = Wave Speed
 𝜔 = Angular Frequency
 A = Amplitude of the
Wave
 Sm / Amplitude of Sound
Wave
 Sm =
𝛥𝑝
𝐵𝑘
 k =
2π𝑓
𝑣
 Sm =
𝛥𝑝
𝐵
2π𝑓
𝑣
 𝛥p = Pressure Amplitude
 B = Bulk Modulus
 f = Frequency
 v = Wave Speed
 Intensity
 I =
1
2
𝜌𝑣𝜔2Sm
2
 𝜌 = Mass Density of
Medium
 v = Wave Speed
 𝜔 = Angular Frequency
 Sm = Amplitude of Sound
Wave

4. Sample Problem 1
The intensity received by a person is 8.62 x 10-10 W/m2, what is the Power delivered from the
source when the person’s eardrum has a radius of 0.8 cm?
I =
𝑃
𝐴
P = I x A
P = 8.62 x 10-10 W/m2 x (π x (0.8 x 10-2 m)2)
P = 1.733 x 10-13 W

5. Sample Problem 2
Known:
T = 0.003 s
𝛥𝛥 p = 30.0 𝜇Pa
Bair = 1.01 x 105 Pa
Formula = Sm =
𝛥𝑝
𝐵
2π𝑓
𝑣
Sm =
𝛥𝑝
𝐵
2π𝑓
𝑣
f =
1
𝑇
=
1
0.003 s
= 333.3 Hz
Sm =
30.0 x 10−6 Pa
1.01 x 105 Pa 2 𝑥 π 𝑥 333.33 𝐻𝑧
343 𝑚/𝑠
Sm = 4.864 x 10-12 m
What is the Amplitude of Sound Wave of a wave with a period of 0.003s and whose
pressure amplitude is 30.0 𝜇Pa? (Bair = 1.01 x 105 Pa)

6. Sample Problem 3
How much power is delivered to a microphone with radius 0.5 mm, by a 2000Hz sound wave
with a displacement amplitude of 4.91 x 10-9 m?
Known:
r = 0.5 mm = 5 x 10-4 m
f = 2000 Hz
Sm = 4.91 x 10-9 m
𝜌𝜌 = 1.2 kg/m3
v = 343 m/s
I =
1
2
𝜌𝑣𝜔2Sm
2
I =
1
2
𝜌𝑣(2π𝑓)2Sm
2
I =
1
2
1.2 kg/m3 (343 m/s)
(2 𝑥 π 𝑥 2000 s−1)2(4.91 x 10-9 m)2
I = 7.835 x 10-7 W/m2
P = I x A
I = 7.835 x 10-7 W/m2
A = π x r2
A = π x (5 x 10-4 m)2
A = 7.854 x 10-7 m2
P = 7.835 x 10-7 W/m2 x 7.854 x 10-7 m2
P = 6.153 x 10-13 W

7. Clicker Question 1:
What is the length of a square plate, that receives an intensity of 9.62 x 10-3 W/m2 that comes
from a speaker with a power of 3.32 x 10 -2 W?
A. 3.451 m
B. 1.858 m
C. 3.194 m
D. 2.896 m
E. 1.7022 m
P = I x A
A =
𝑃
𝐼
A =
3.32 x 10 −2 W
9.62 x 10−3 W/m2
A = 3.451 m2
Length of a square = 𝐴
Length = 3.451 m2 = 1.858 m

8. Clicker Question 2:
What is the Power exerted per unit area on a basketball with a radius of 15cm, from a 500 Hz
sound wave that has an Amplitude of 3.0 x 10-5 m?
A. 1.342 x 10-4 W/m2
B. 1.342 x 10-3 W/m2
C. 8.412 x 104 W/m2
D. 4.747 x 10-3 W/m2
E. 4.4747 x 10-4 W/m2
Known:
I =
𝑃
𝐴
P =
1
2
𝜇𝑣𝜔2A2
𝜇 = 8.81 x 10-5 kg/m
A = 3.0 x 10-5 m
f = 500 Hz
R = 0.15 m
P =
1
2
𝜇𝑣𝜔2A2
P =
1
2
𝜇𝑣(2π𝑓)2A2
P =
1
2
(8.81 x 10-5 kg/m) (343 m/s) (2 x π x 500Hz)2
(3.0 x 10-5 m)2
P = 1.342 x 10-4 W
A = 4πr2 = 4 x π x 0.152 = 0.2827m2
I =
𝑃
𝐴
=
1.342 x 10−4 W
0.2827 m2 = 4.747 x 10-4 W/m2