Physics 101 Learning Object 04

1. Physics 101 Learning Object 4 - Sound and Sound Intensity
Bruno Freitas, 11476141, Section L2H
Question:
Scientists have found a material that seems to be made of a pure unknown element. In
order to figure out which element this material is made of, the scientists observed the
speed that sound passes through the medium, which provided a speed of 1450m/s
a. Given the table of densities of the possible elements, and an observed Bulk
modulus of 2.85 x 1010 Pa, what is the element of the medium?
b. In this experiment, scientists used a 1300 Hz sound wave, which produced 1.2 x
10-12 W of power. Determine the amplitude of the sound wave through this
medium (the average radius of a human eardrum is 0.75cm)
c. How many decibels were produced by this sound? (the reference sound intensity
Io = 10-12 W/m2)

2. Answers:
Scientists have found a material that seems to be made of a pure unknown element. In
order to figure out which element this material is made of, the scientists observed the
speed that sound passes through the medium, which provided a speed of 1450m/s
a. Given the table of densities of the possible elements, and an observed Bulk
modulus of 2.85 x 1010 Pa, what is the element of the medium?
To find the answer of this question, we must first use the equation:
v = √
𝛽
𝜌
By using the values given to us, we are able to solve for density:
1450m/s = √
2.85 ∙ 1010
𝑃𝑎
𝜌
By using algebra, we can solve for density:
𝜌 =
2.85 ∙ 1010
14502
𝜌 = 13555.29 kg/m3
Therefore, by looking at the table given to us, we can determine the element of the
medium is Mercury.
b. In this experiment, scientists used a 1300 Hz sound wave, which produced 1.2 x
10-12 W of power. Determine the amplitude of the sound wave through this
medium (the average radius of a human eardrum is 0.75cm)
In this question there are two formulas we must use to find the answer,

3. I =
1
2
𝜌𝜔2
𝑠 𝑚
2
and P = IA
i) To solve for Intensity:
P = IA
1.2 x 10-12 W = I x π x 0.0075 m2
By isolating “I” we determine
I = 6.7906 x 10-9 W/m2
ii) To solve for sm:
I =
1
2
𝜌𝜔2
𝑠 𝑚
2
6.7906 x 10-9 W/m2 =
1
2
13555.29 ∙ 2𝜋 ∙ (1300𝑠−1
)2
∙ 𝑠 𝑚
2
By using algebra and isolating sm, we get:
sm = √ 2 ∙ (6.7906 ∙ 10−9
)
13555.29 ∙ 2𝜋 ∙ 13002
sm = 3.07 x 10-10m
Therefore, the maximum displacement (Amplitude) of the medium is 3.07 x 10-10m.
c. How many decibels were produced by this sound? (the reference sound intensity
Io = 10-12 W/m2)
With this last question, in order to find how many decibels are being produced by the
sound we must use the equation:
𝛽(𝐼) = 0 𝑑𝐵 + 10𝑙𝑜𝑔10(
𝐼
𝐼 𝑜
)
By using the given reference sound intensity, and the intensity we have calculated in the
previous question, we can determine dB:
𝛽(𝐼) = 0 𝑑𝐵 + 10𝑙𝑜𝑔10(
6.7906 ∙ 10−9
𝑊/𝑚2
1 ∙ 10−12
𝑊/𝑚2
)

4. 𝛽(𝐼) = 38. 32 dB
Therefore, the decibels produced by the sound were approximately 38 decibels.