A fixed string in the third harmonic

[TFM]

[SOLVED] A fixed string in the third harmonic

Homework Statement

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 195m/s and a frequency of 230hz The amplitude of the standing wave at an antinode is 0.370cm.

Calculate the amplitude at point on the string a distance of 20.0cm from the left-hand end of the string.

Homework Equations

The Attempt at a Solution

I'm not sure how to go about this question? Any tips?

TFM

 

[Oerg]

Firstly, You have to be asking yourelf, what does it mean for an oscillation to be in the third harmonic?

 

[TFM]

The formula for harmonics is:

$$f_n = n\frac{v}{2L}$$

and the third harmonic is:

$$f_3 = 3\frac{v}{2L}$$

The third Harmionc al,os has three antinodes, and 1.5 wavelengths

TFM

 

[Oerg]

Now, using v=f\lambda, you can calculate the wavelength.

The question asks for 20cm from a node. Since you know the amplitude at the antinode, can you now calculate the amplitude at a given point in the wave

 

[TFM]

I have calculated the wavelength to be 1.18m

This implies the length of string is 1.77m,

TFM

 

[TFM]

I have now caluclated the nodes to be at:

0, 0.59, 1.18, 1.77

and the anti-nodes at

0.295, 0.885, 1.475

TFM

 

[TFM]

Have I correctly worked out what to do now?

using:

$$y(x,t) = A_S_Wsin(k_n x)sin(\omega _n t)$$

where:

$$A_S_W = 0.0037 cm$$

$$k_n = 2\pi / 1.18$$

[tex] \omega_n = 2\pi / (1/frequency) [\tex]

giving:

$$y(x,t) = 0.0037*sin((2\pi / 1.18) 0.2)sin(2\pi / (1/230) t)$$

I just need what t is, I think. Is this right?

TFM

 

[Vuldoraq]

You should be using $$v=f*\lambda$$ to find the wavelength.

 

[TFM]

Rearranges to give wavelength = v/f

gives wavelength of 0.848m

nodes are at 0, 0.424m, 0.848m and 1.271m

antinodes 0.212m, 0.635m, 1.06m

length of string is 1.271m

$$k_n = \frac{2\pi}{0.848}$$

am I using the correct formula for the amplitude

$$y(x,t) = A_S_W sin(k_n x)sin(\omega _n t)$$

and if so, how do I work out the time, t

TFM

 

[Vuldoraq]

Hi,

Your equation looks good, except for your value of the wavelength. If you recalculate your wavelength using $$v=f*\lambda$$ and then put this into your equation you will be good to go.

In this question remember that you are only considering the position of the particle and in such an instance you can make a conclusion about the time dependence.

 

[Vuldoraq]

Sorry, I didn't see that you had corrected the wavelength above. Yeah your formula is the right to use in this case.

 

[TFM]

Would it be best to presume that the time = 0?

TFM

 

[Vuldoraq]

No, if you do that the equation goes to zero and you will get 0 as the amplitude.

The amplitude at any position on the string does not depend on the time (if you were interested in the vertical displacement of the particle then the time would be relevant). The amplitude of a particle on a transverse wave depends only on the position of that particle. (Remember the amplitude doesn't change with time unless the wave is being damped, which it isn't in this case.)

 

[TFM]

What would be a good time to use?

TFM

 

[Vuldoraq]

You just ignore the time part of the equation ie $$sin(\omega*t)$$ to give,

$$y(x)=A_{SW}sin(k_{n}x)$$ which should be the amplitude.

 

[TFM]

I get an answer of 0.0036854, which seems too close to the antinodes 0.0037?

TFM

 

[Vuldoraq]

That should be the right answer. Don't forget that the first antinode occurs at a distance of 21.2 cm from the left of the rope, which is very close to the point under observation, thus their amplitudes will be quite close.

 

[Vuldoraq]

Do you mind me asking what university you go to? It's odd that you seem to do all the same problems that I'm set. I can't work out how to do the second part of that question you are doing.

 

[TFM]

The second parts okay (I am presuming you are referring tgo the time to get from its largest upward disp. to its largest downwards disp.), you have to think when it is at its highest point and when it is at it's lowest point. try drawing adiagram can help.

TFM

 

[TFM]

I am now stuck on parts C and D, finding the maximum transverse velocity and the maximum transverse acceleration. The key points, I feel, though is that the velocity is greatest when the amplitude is 0 (when it crosses the normal position) and the acceleration is greatest when it reaches maximum/minimum amplitude.

Any suggestions?

TFM

 

[Vuldoraq]

For C and D, just recall that velocity is the first time derivative and acceleration is the second time derivative. Then recall that cos(angle) and sin(angle) have a maximum value of 1.

 

[Vuldoraq]

So for C and D you don't need to know the time you just say that the time part has a maximum value of 1.

I don't get what you mean by thinking about when the particle is at it's highest and lowest points. I only have one attempt left on MP so I don't want to guess again. Is the answer something like,

$$(f/2)^{-1}$$

 

[TFM]

I used 1/2 *(1/f), which I think is the same as what you put - this is from the fact that it goes from its maximam point to its minimum point in half a period.

work out your answer and post it to check first.

I had a feeling it would end up using them equations.

TFM

 

[TFM]

Original equation:

$$y(x,t) = A_S_W sin(k_n x) sin(\omega t)$$

since we are ignoring t,

$$y(x,t) = A_S_W sin(k_n x) sin(\omega )$$

so velocity:

$$v(x,t) = A_S_W x cos(k_n x)$$

and acceleration:

$$v(x,t) = -A_S_W x^2 sin(k_n x)$$

Do these look right?

TFM

 

[Vuldoraq]

Not quite. You've differentiated with respect to x what you want is the time derivative. In this part we aren't ignoring the time bit we just say that it has a max value of 1.

this is the derivative you need to do;

$$\delta y(x,t)/\delta t=\delta A_{SW}sin(k_{n}x)sin(\omega_{n}t)/\delta t$$

 

[Vuldoraq]

Have you done partial differentiation? You should get for the velocity part,

$$v(x,t)=\omega_{n}A_{SW}sin(k_{n}x)cos(\omega_{n}t)$$

Which has a maximum value when $$cos(\omega_{n}t)=1$$

So this has maximum values when $$t=n*(1/f)$$ where n=(1,2,3,...),

 

[TFM]

Should it be:

$$v(x,t) = tA_S_W sin(k_n x)cos(\omega_n t)$$

TFM

Eidt, didn't see that last post, sorry, yeah, silly mistake on my part.

 

[Vuldoraq]

Actually n=0 is also possible.

My value of f is 235Hz and I get my period to be 8.51*10^-3 s

 

[Vuldoraq]

The t doesn't come out in front, by the chain rule;

$$y(x,t)=sin(\omega_{n}t) let \omega_{n}t=u then$$

$$dsin(u)/du=cos(u) and d\omega_{n}t/dt=\omega_{n} which is du/dt$$

$$since d(sin(\omega_{n}t)/dt=(du/dt)*(dy(u)/du)$$

$$d(sin(\omega_{n}t)/dt=cos(u)\omega_{n}$$

$$which is cos((\omega_{n}t) )\omega_{n}$$

Edit: My bad I don't think this makes it any clearer, I thought latex would recognise the spaces between words!

 

[TFM]

You have to end each latex to put written explanations between lines.

You may want to check your periods, period = 1/f, which gives your period to be 0.004.


Using:

$$v(x,t)=\omega_{n}A_{SW}sin(k_{n}x)cos(\omega_{n}t)$$

omega = 2pi/period

A = 0.0037

cos(w t) = cos(0)

should the sin(k x) also be sin (pi/2)?

TFM

 

[TFM]

Using:

$$v(x,t) = A_S_W \omega sin(\frac{\pi }{2}) cos(0)$$

gives me: 5.33

then

$$a(x,t) = A_S_W \omega ^2 sin(\frac{\pi }{2}) sin(\frac{\pi }{2})$$

gives me: 7700

Both of which are right. I disagree with mastering physics - if you diffentiate cos, it gives you minus sin:

$$a(x,t) = -A_S_W \omega ^2 sin(\frac{\pi }{2}) sin(\frac{\pi }{2})$$ giving -7700, which it said was wrong?

Many Thanks,

TFM

 

[Vuldoraq]

Indeed you do get minus sin, but -7700 would be the maximum deceleration, and it asks for the maximum acceleration. At least that's what I tell myself.

Thanks for your help with the period, I think I was having a mental bock.

 

[TFM]

Have you got the right answer yet for the time, as I have left the thread marked unsolved until you have finished.

I think I was having a mental bock.

Don't worry, I think it happens to all of us from time to time!

TFM

 

[Vuldoraq]

Yeah, I got in the end. Thanks for that .

By the way are you still working on the 'instantaneous power in a standing wave' problem? Sketching the graphs is quite tricky.