### Passage A student conducts three experiments to study the fundamental frequency, f (in… — ACT

### Passage

A student conducts three experiments to study the fundamental frequency, $f$ (in hertz, $\text{Hz}$ ), of a vibrating string on a sonometer.

In Experiment 1, the student varies the length of the string, $L$ (in meters, $\text{m}$ ), while keeping the tension, $T$ (in newtons, $\text{N}$ ), and the linear mass density, $\mu$ (in grams per meter, $\text{g/m}$ ), constant.

In Experiment 2, the student varies the tension, $T$ , while keeping the length ( $L = 0.50\ \text{m}$ ) and linear mass density ( $\mu = 2.0\ \text{g/m}$ ) constant.

In Experiment 3, the student varies the linear mass density, $\mu$ , by using different strings while keeping the length ( $L = 0.50\ \text{m}$ ) and tension ( $T = 100\ \text{N}$ ) constant.

The results of the three experiments are recorded in the tables below:

Table 1 (Experiment 1)

Trial	Length $L$ ( $\text{m}$ )	Frequency $f$ ( $\text{Hz}$ )
1	$0.25$	$440$
2	$0.50$	$220$
3	$1.00$	$110$

Table 2 (Experiment 2)

Trial	Tension $T$ ( $\text{N}$ )	Frequency $f$ ( $\text{Hz}$ )
4	$25$	$110$
5	$100$	$220$
6	$400$	$440$

Table 3 (Experiment 3)

Trial	Linear mass density $\mu$ ( $\text{g/m}$ )	Frequency $f$ ( $\text{Hz}$ )
7	$0.5$	$440$
8	$2.0$	$220$
9	$8.0$	$110$

Based on the tables, match each physical relationship to the equation that correctly describes the proportionality and fits the experimental data.

Frequency ( $f$ ) as a function of string length ( $L$ ) when $T$ and $\mu$ are constant $f = \frac{110}{L}$
Frequency ( $f$ ) as a function of tension ( $T$ ) when $L$ and $\mu$ are constant $f = 22\sqrt{T}$
The square of the frequency ( $f^2$ ) as a function of linear mass density ( $\mu$ ) when $L$ and $T$ are constant $f^2 = \frac{96,800}{\mu}$
Frequency ( $f$ ) as a function of linear mass density ( $\mu$ ) when $L$ and $T$ are constant $f = \frac{220\sqrt{2}}{\sqrt{\mu}}$

Answer

Frequency as a function of length is

f = \frac{110}{L}

; frequency as a function of tension is

f = 22\sqrt{T}

; the square of frequency as a function of linear mass density is

f^2 = \frac{96,800}{\mu}

; and frequency as a function of linear mass density is

f = \frac{220\sqrt{2}}{\sqrt{\mu}}

The correct pairings are established by identifying the constant product or ratio for each set of experimental trials. For Experiment 1, the product

f \cdot L

is constant at

110

, showing an inverse relationship

f = \frac{110}{L}

. For Experiment 2, the ratio

\frac{f}{\sqrt{T}}

is constant at

22

, showing a direct relationship to the square root,

f = 22\sqrt{T}

. For Experiment 3, the product

f^2 \cdot \mu

is constant at

96,800

, showing that

f^2

is inversely proportional to

\mu

, which simplifies to

f = \frac{220\sqrt{2}}{\sqrt{\mu}}

Step-by-Step Solution

Analyze the relationship between frequency (

f

) and string length (

L

) using Table 1.

Doubling

L

from

0.25\ \text{m}

0.50\ \text{m}

halves

f

from

440\ \text{Hz}

220\ \text{Hz}

, indicating that

f

is inversely proportional to

L

(

f = \frac{k_1}{L}

). Solving for the constant gives

k_1 = f \cdot L = 220 \times 0.50 = 110

. Thus,

f = \frac{110}{L}

To determine the equation for frequency as a function of length under constant tension and linear mass density.

Analyze the relationship between frequency (

f

) and tension (

T

) using Table 2.

Quadrupling

T

from

25\ \text{N}

100\ \text{N}

doubles

f

from

110\ \text{Hz}

220\ \text{Hz}

, indicating that

f

is directly proportional to the square root of tension (

f = k_2\sqrt{T}

). Solving for the constant gives

k_2 = \frac{f}{\sqrt{T}} = \frac{220}{\sqrt{100}} = 22

. Thus,

f = 22\sqrt{T}

To determine the equation for frequency as a function of tension under constant length and linear mass density.

Analyze the relationship between the square of the frequency (

f^2

) and linear mass density (

\mu

) using Table 3.

Quadrupling

\mu

from

0.5\ \text{g/m}

2.0\ \text{g/m}

halves

f

from

440\ \text{Hz}

220\ \text{Hz}

, meaning that

f^2

is quartered from

193,600\ \text{Hz}^2

48,400\ \text{Hz}^2

. This shows that

f^2

is inversely proportional to

\mu

(

f^2 = \frac{k_3}{\mu}

). Solving for the constant gives

k_3 = f^2 \cdot \mu = 220^2 \times 2.0 = 96,800

. Thus,

f^2 = \frac{96,800}{\mu}

To determine the equation for the square of the frequency as a function of linear mass density.

Derive the direct relationship between frequency (

f

) and linear mass density (

\mu

) using the equation from Step 3.

Taking the square root of

f^2 = \frac{96,800}{\mu}

yields

f = \frac{\sqrt{96,800}}{\sqrt{\mu}} = \frac{220\sqrt{2}}{\sqrt{\mu}}

To express the frequency as a function of the square root of linear mass density.

Key Concept

Direct and inverse proportionality in physical systems, including relationships involving roots and powers of variables.

Estimated Time:2m 0s

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