Standing Wave Analysis on a Stretched Wire with Fixed Endpoints

[SirRoland]

Homework Statement

A wire with mass 35.0 is stretched so that its ends are tied down at points a distance 83.0 apart. The wire vibrates in its fundamental mode with frequency 64.0 and with an amplitude at the antinodes of 0.270 .

it asks for: Tension
Speed of propagation
maxmimum transverse velocity and
maximum acceleration

Homework Equations

Wave equation Asin(kx-wt)

I figured I was suppose to put the equation into this form first, however I was really unsure because I have never done a problem like this before.

The Attempt at a Solution

.270cmSin(7.57x-402.12t)
.270 is given
7.57 i took 2pi/.83 : here I am assuming .83m (83 cm given) is the wavelength, as I did not know what else it could be.
402.12 i have from frequency * 2pi.

I am not sure at all if this is a step in the right direction, any help is really appreciated :)

 

[danielakkerma]

Hi,
Lets resolve the problem one step at a time:
Think about Newtonian mechanics, for the part regarding the tension.
Remember that the force is $$\vec{F}=m \cdot \vec{a}$$
Think about what you need to get that..
The same applies to the velocity, and acceleration.
Recall(and this is quite a bulky hint) the equation you've been given is that of the translation, i.e:
$$x=Asin(\omega t - kx)$$

Daniel

 

[Je m'appelle]

"A wire with mass 35.0 is stretched so that its ends are tied down at points a distance 83.0 apart."

As the endpoints of the wire are fixed the wave formed on this wire will be a standing wave.

"The wire vibrates in its fundamental mode with frequency 64.0 and with an amplitude at the antinodes of 0.270."

Do you know what's the concept of fundamental frequency mode of vibration in a standing wave? Do you know the definition of antinodes?

The general vibration equation for a standing wave:

$$y_n(x,t) = b_n sin(k_n x) cos(\omega_n t + \delta_n)$$

Where

$$k_n = \frac{n \pi}{L}$$

$$\omega_n = k_n v = \frac{n \pi v}{L}$$

And

$$\lambda_n = \frac{2 \pi}{k_n} = \frac{2L}{n}$$

$$\nu_n = \frac{\omega_n}{2 \pi} = n \frac{v}{2L}$$

$$v = \sqrt{\frac{T}{\mu}}$$

For a fundamental vibration, $$n = 1$$.

Glossary:

$$b_n$$ is the amplitude at the nth mode of vibration.

$$\delta_n$$ is the phase angle (in case there is one) at the nth mode of vibration.

$$k_n$$ is the wavenumber at the nth mode of vibration.

$$\omega_n$$ is the angular frequency at the nth mode of vibration.

$$\lambda_n$$ is the wave length at the nth mode of vibration.

$$\nu_n$$ is the frequency at the nth mode of vibration.

$$v$$ is the velocity of propagation of the wave in the wire.

$$T$$ is the tension of the wire.

$$\mu$$ is the linear density of the wire.

The speed of propagation should be the speed of the wave $$v$$.

The maximum transverse velocity can be found through $$\frac{d}{dt}y_n(x,t)$$

The maximum transverse acceleration can be found through $$\frac{d^2}{dt^2}y_n(x,t)$$

Note that in order to find the maximum value for the transverse velocity and acceleration you have to analyze when $$\frac{d}{dt}y_n(x,t)$$ and $$\frac{d^2}{dt^2}y_n(x,t)$$ achieve maximum value respectively.

All of this should be in your textbook.

Now use them according to the values given.

If there's anything you didn't understand feel free to ask.