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Help with stationary waves on strings ?

Hi, I'm attempting questions on this topic and I'm stuck i've looked on the internet, used my textbook, and my brain, I have tried really hard but I have no idea what to do please help, Thanks :smile:

2) The first harmonic frequency of vibration of a stretched wire is inversely proportional to the length of the wire. For the wire in 1 at the same tension, Calculate the length of the wire to product a frequency of A) 512hz B) 384hz ??

3) The tension in the wire in Q1 was 40N. Calculate A) the mass per unit length of the wire ? ( I've done this and the answer is 2.4^-4 kgm^-1),

B) the diameter of the wire if its density (mass per unit volume) was 7800 kg m^-3 ??

I'm assuming you need question 1 too which ive already answered which is below:

1) A stretched wire of length 0.80m vibrates at its first harmonic with a frequency of 256hz . Calculate A) the wavelength of the progressive waves on the wire (answer 1.6m)
B) the speed of the progressive waves on the wire (410 ms^-1)

Sorry for the long winded questions but i really need help and don't know where to ask, Thank you so much I really appreciate it :smile:
Original post by Cinna21
Hi, I'm attempting questions on this topic and I'm stuck i've looked on the internet, used my textbook, and my brain, I have tried really hard but I have no idea what to do please help, Thanks :smile:

1) A stretched wire of length 0.80m vibrates at its first harmonic with a frequency of 256hz . Calculate A) the wavelength of the progressive waves on the wire (answer 1.6m)
B) the speed of the progressive waves on the wire (410 ms^-1)

2) The first harmonic frequency of vibration of a stretched wire is inversely proportional to the length of the wire. For the wire in 1 at the same tension, Calculate the length of the wire to product a frequency of A) 512hz B) 384hz ??


If we let l l denote the length of the wire and f f denote the frequency, then what question 2 tells us is that

l=kf l = \frac{k}{f} , where k k is a constant of proportionality.

If we knew the value of our constant, we could just substitute our frequencies into that formula and compute the tension.

However, because this is the same wire at the same tension, we can say that the constant of proportionality will be same for all the frequencies and, using part 1 knowledge, we can work out what exactly this constant is.

From part 1, you know that a wire of length 0.8m produces its first harmonic at a frequency of 256Hz.

So do you think you can work out the constant of proportionality and go from here?

Original post by Cinna21
3) The tension in the wire in Q1 was 40N. Calculate A) the mass per unit length of the wire ? ( I've done this and the answer is 2.4^-4 kgm^-1),

B) the diameter of the wire if its density (mass per unit volume) was 7800 kg m^-3 ?

Sorry for the long winded questions but i really need help and don't know where to ask, Thank you so much I really appreciate it :smile:


For part b, you know that density = mass/volume. I'm pretty sure that you can assume that the cross section of the wire is circular, and hence if you work out the volume, you can rearrange for diameter easily.

Hope that helps :smile:
Reply 2
Original post by kingaaran
If we let l l denote the length of the wire and f f denote the frequency, then what question 2 tells us is that

l=kf l = \frac{k}{f} , where k k is a constant of proportionality.

If we knew the value of our constant, we could just substitute our frequencies into that formula and compute the tension.

However, because this is the same wire at the same tension, we can say that the constant of proportionality will be same for all the frequencies and, using part 1 knowledge, we can work out what exactly this constant is.

From part 1, you know that a wire of length 0.8m produces its first harmonic at a frequency of 256Hz.

So do you think you can work out the constant of proportionality and go from here?



For part b, you know that density = mass/volume. I'm pretty sure that you can assume that the cross section of the wire is circular, and hence if you work out the volume, you can rearrange for diameter easily.

Hope that helps :smile:





Hi thanks soo much for your help :biggrin: I answered 3B thanks to your help but I still don't know how to answer 2a & 2b everything you said seems alien to me, could you answer it, thank you so much and I appreciate your time :smile:

Quick Question : For 3B which I answered, I got my answer in meters but the answer to that question is in mm, when I check answers in an exam how would I know I need to change it to mm or should I keep everything in meters.

Thanks again :smile:
Original post by Cinna21
Hi thanks soo much for your help :biggrin: I answered 3B thanks to your help but I still don't know how to answer 2a & 2b everything you said seems alien to me, could you answer it, thank you so much and I appreciate your time :smile:

Quick Question : For 3B which I answered, I got my answer in meters but the answer to that question is in mm, when I check answers in an exam how would I know I need to change it to mm or should I keep everything in meters.

Thanks again :smile:


If two quantities are inversely proportional, then they follow the relationship A=kB A = \frac{k}{B} , where A and B are our quantities and k is a constant. This is GCSE maths and you should look it up if you're unsure.

Now, we are told that the frequency of the wire is inversely proportional to its length. So we can say

f=kl f = \frac{k}{l} or l=kf l = \frac{k}{f} - it's the same thing.

From part 1, we are told that if we have a length of 0.8m, we get a frequency of 256Hz. Substituting that in, we get:

256=k0.8k=204.8 256 = \frac{k}{0.8} \rightarrow k = 204.8

So our constant (something that stays the same) is 204.8 and our relationship is now:

l=204.8f l = \frac{204.8}{f}

So if we want a frequency of 512Hz,

l=204.8512 l = \frac{204.8}{512} , the length of our wire must be 0.40m.

Doing a similar calculation for 384Hz, we find the length of our wire must be 0.53m (3sf).

Does that help?
Reply 4
Original post by kingaaran
If two quantities are inversely proportional, then they follow the relationship A=kB A = \frac{k}{B} , where A and B are our quantities and k is a constant. This is GCSE maths and you should look it up if you're unsure.

Now, we are told that the frequency of the wire is inversely proportional to its length. So we can say

f=kl f = \frac{k}{l} or l=kf l = \frac{k}{f} - it's the same thing.

From part 1, we are told that if we have a length of 0.8m, we get a frequency of 256Hz. Substituting that in, we get:

256=k0.8k=204.8 256 = \frac{k}{0.8} \rightarrow k = 204.8

So our constant (something that stays the same) is 204.8 and our relationship is now:

l=204.8f l = \frac{204.8}{f}

So if we want a frequency of 512Hz,

l=204.8512 l = \frac{204.8}{512} , the length of our wire must be 0.40m.

Doing a similar calculation for 384Hz, we find the length of our wire must be 0.53m (3sf).

Does that help?


Thank you so much, I really appreciate your time, I need to go over this more but I've answered the questions, Thank you so much & Merry Christmas :P
Original post by penguinboi
I know this is an old thread but can someone help me with part b?


Assuming that you are referring to this question:
"1) A stretched wire of length 0.80m vibrates at its first harmonic with a frequency of 256hz . Calculate A) the wavelength of the progressive waves on the wire (answer 1.6m)

B) the speed of the progressive waves on the wire (410 ms^-1)"
Distance between 2 adjacent nodes is half a wavelength
The wavelength is 2 x 0.8 = 1.6 m

speed = frequency x wavelength = 256 x 1.6 = 410 m/s
Original post by penguinboi
I need help with this question: B) the diameter of the wire if its density (mass per unit volume) was 7800 kg m^-3 ??

But you have to use information from Q1


The mass per unit length (linear mass density) is given by
μ=ρA \mu = \rho A
hence
μ=ρπd24 \mu = \rho \pi \dfrac{d^2}{4}
from which we solve for diameter d, using the mass per unit length from part B and the density given
d = 2.0 x 10^-4 m
Reply 7
what's p?
Reply 8
Density

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