QuestionSinusoidal waves 5.00 cm in amplitude are to be transmitted along a string that has a linear mass density of $4.00 \times 10^{-2} \mathrm{~kg} / \mathrm{m}$ . The source can dellver a maximum power of 280 W , and the string is under a tension of 90 N . What is the highest frequency $f$ at which the source can operate?
Part 1 of 5 - Conceptualize Increasing the source frequency increases the power carried by the wave. A power of 280 W will possibly occur at a frequency on the order of 100 Hz .
Part 2 of 5 - Categorize We will use the expression for power carried by a wave on a string.
Part 3 of 5 - Analyze The wave speed is given by $\begin{array}{l} v=\sqrt{\frac{T}{\mu}} \\ \sqrt{\frac{\square \mathrm{~N}}{\square \mathrm{~m} / \mathrm{s}}} \\ =\square 0^{-2} \mathrm{~kg} / \mathrm{m} \end{array}$

Studdy Solution

STEP 1

1. The amplitude of the sinusoidal waves is 5.00 cm.
2. The linear mass density of the string is $4.00 \times 10^{-2} \, \mathrm{kg/m}$ .
3. The maximum power delivered by the source is 280 W.
4. The tension in the string is 90 N.
5. We are tasked with finding the highest frequency $f$ at which the source can operate.

STEP 2

1. Conceptualize the relationship between frequency and power.
2. Categorize the problem using the expression for power carried by a wave on a string.
3. Analyze the wave speed.
4. Derive the expression for power in terms of frequency.
5. Solve for the highest frequency.

STEP 3

Conceptualize the relationship between frequency and power. Increasing the frequency of the source increases the power carried by the wave. The maximum power of 280 W is expected to occur at a frequency on the order of 100 Hz.

STEP 4

Categorize the problem using the expression for power carried by a wave on a string. The power $P$ carried by a wave on a string is given by:
$P = \frac{1}{2} \mu v \omega^2 A^2$
where: - $\mu$ is the linear mass density, - $v$ is the wave speed, - $\omega$ is the angular frequency, - $A$ is the amplitude.

STEP 5

Analyze the wave speed. The wave speed $v$ on a string under tension $T$ is given by:
$v = \sqrt{\frac{T}{\mu}}$
Substitute the given values:
$v = \sqrt{\frac{90 \, \mathrm{N}}{4.00 \times 10^{-2} \, \mathrm{kg/m}}}$
Calculate $v$ :
$v = \sqrt{\frac{90}{0.04}}$ $v = \sqrt{2250}$ $v = 47.43 \, \mathrm{m/s}$

STEP 6

Derive the expression for power in terms of frequency. The angular frequency $\omega$ is related to the frequency $f$ by $\omega = 2\pi f$ . Substitute $\omega = 2\pi f$ into the power expression:
$P = \frac{1}{2} \mu v (2\pi f)^2 A^2$
Simplify:
$P = 2 \pi^2 \mu v f^2 A^2$
Solve for $f$ :
$f^2 = \frac{P}{2 \pi^2 \mu v A^2}$

STEP 7

Solve for the highest frequency. Substitute the given values into the expression for $f^2$ :
$f^2 = \frac{280}{2 \pi^2 (4.00 \times 10^{-2}) (47.43) (0.05)^2}$
Calculate $f^2$ :
$f^2 = \frac{280}{2 \pi^2 \times 0.04 \times 47.43 \times 0.0025}$
$f^2 = \frac{280}{0.023561944}$
$f^2 = 11879.63$
Calculate $f$ :
$f = \sqrt{11879.63}$
$f \approx 109 \, \mathrm{Hz}$
The highest frequency at which the source can operate is approximately:
$\boxed{109 \, \mathrm{Hz}}$

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