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Mechanical Waves On a String - Speed, Amplitude, and Power

by Nickg140143
Tags: mechanical waves, power, rope, speed
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Mar11-12, 08:31 PM
P: 30
1. The problem statement, all variables and given/known data
A string of mass 38.5g and length 5.60m is secured so that it is under tension of 220N. A wave with frequency 178 Hz travels on the string. Find the speed of the wave and the amplitude of the wave if it transmits power of 140 Watts.

The Given answers are: 179 m/s and 1.35 cm

2. Relevant equations
I believe these are the main equations that I can use to solve these problems

speed of wave
[itex]v=\sqrt{\frac{Tension}{\mu}},\mu = \frac{Mass}{Length}[/itex]

Power (from my notes)

[itex]P=\frac{\mu \times v \times \omega^2 \times A^2}{2},\omega = 2\pi f[/itex]

Power (from my book, think this is average power)

[itex]P=\frac{\sqrt{\mu \times F} \times \omega^2 \times A^2}{2},\omega = 2\pi f[/itex]

Are my notes correct? Are these equations for the same amount of power?

3. The attempt at a solution

Well, I'm given mass, length and tension of the string, and If my understanding is somewhat correct, the speed at which a wave moves through a medium is dependent only on the properties of the medium itself, and for this string, these properties are the mass, length, and tension

[itex] mass(M) = 38.5 g(grams) \longrightarrow .385 kg(kilograms)[/itex]
[itex] tension(T) = 220N(\frac{kgm}{s^2})[/itex]
[itex] length(L)=5.60m(meters)[/itex]

With this information, I can use my formula for velocity of the wave to find the speed of the wave.

I'll calculate [itex]\mu[/itex]
[itex]\mu = \frac{M}{L} \longrightarrow \mu = \frac{.385 kg}{5.60 m}=0.069\frac{kg}{m}[/itex]

Now [itex]v[/itex]
[itex]v=\sqrt{\frac{T}{\mu}}\longrightarrow v=\sqrt{\frac{220\frac{kgm}{s^2}}{0.069\frac{kg}{m}}}=\boxed{56.5\frac{ m}{s}}[/itex]

Since I now have the velocity [itex]v[/itex], I should use the equation for power [itex]P[/itex], where [itex]P = 140 watts(\frac{kgm^2}{s^3})[/itex] and solve for A

[itex]P=\frac{\mu \times v \times \omega^2 \times A^2}{2},\omega = 2\pi f[/itex]

but before I can use this, I need [itex]\omega[/itex], which is [itex]\omega = 2\pi \times frequencey(f)[/itex]

[itex]\omega = 2\pi(rads) \times 178 \frac{1}{s}=1118.41\frac{1}{s}[/itex]

now that I have [itex]\omega[/itex], I can solve for A

[itex]140\frac{kgm^2}{s^3}=\frac{(0.069\frac{kg}{m}) \times (56.5\frac{m}{s}) \times(1118.41\frac{1}{s})^2 \times A^2}{2}[/itex]

[itex]\frac{280\frac{kgm^2}{s^3}}{(0.069\frac{kg}{m}) \times (56.5\frac{m}{s}) \times(1118.41\frac{1}{s})^2}=A^2[/itex]

[itex]A^2=5.74\times 10^{-5}m[/itex] or [itex]0.0000574 m \longrightarrow \boxed{A=.0076m \rightarrow .76cm}[/itex]

But as you can see, these answers do not correspond with those that are given. There are plenty of concepts that I'm struggling with here in physics, so I want to make sure I'm not straying too far off the path here.

any help would be greatly appreciated.
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Mar14-12, 01:51 PM
P: 30
Found my problem....


This mass should actually be


Looks like I simply messed up on a conversion.

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