Second harmonic waves HELP!

Attachment not found

Can someone please explain how I could answer this question? I've attached the question above

What I have tried already is:

That the frequency of a wave can be given as follows:

f = 1/2L * sqrt (T/m)

where L = length of oscillating string , T = tension and m = mass per unit length.

I then know that 2L is the same as lambda (wavelength), so I rearranged and got this now:

lambda = 1/f * sqrt (T/m)

I think that since f and T are both constants I can say that 'lambda is proportional to 1/sqrt (m).

But I'm stuck on this part and don't where to go and move on from there.

Any help would be appreciated. Thanks

(edited 5 years ago)

Original post by Yatayyat

Attachment not found

Can someone please explain how I could answer this question? I've attached the question above

What I have tried already is:

That the frequency of a wave can be given as follows:

f = 1/2L * sqrt (T/m)

where L = length of oscillating string , T = tension and m = mass per unit length.

I then know that 2L is the same as lambda (wavelength), so I rearranged and got this now:

lambda = 1/f * sqrt (T/m)

I think that since f and T are both constants I can say that 'lambda is proportional to 1/sqrt (m).

But I'm stuck on this part and don't where to go and move on from there.

Any help would be appreciated. Thanks

You may want to repost the attachment. I cannot see the question in the attachment.

Reply 2

Yatayyat

OP

14

Original post by Eimmanuel

You may want to repost the attachment. I cannot see the question in the attachment.

I have reattached it now.

Original post by Yatayyat

I have reattached it now.

Sorry just saw it. :smile:

Original post by Yatayyat

Attachment not found

Can someone please explain how I could answer this question? I've attached the question above

What I have tried already is:

That the frequency of a wave can be given as follows:

f = 1/2L * sqrt (T/m)

where L = length of oscillating string , T = tension and m = mass per unit length.

I then know that 2L is the same as lambda (wavelength), so I rearranged and got this now:

lambda = 1/f * sqrt (T/m)

I think that since f and T are both constants I can say that 'lambda is proportional to 1/sqrt (m).

But I'm stuck on this part and don't where to go and move on from there.

Any help would be appreciated. Thanks

First of all, you should write the frequency of the normal modes of the standing wave is

f_n = \dfrac{n}{2L} \sqrt{\dfrac{T}{\mu}}

where n= 1, 2, 3…

if you want to write the general equation for the frequency instead of

Original post by Yatayyat

That the frequency of a wave can be given as follows:

f = 1/2L * sqrt (T/m)

where L = length of oscillating string , T = tension and m = mass per unit length.
Because the frequency that you stated is the fundamental frequency.

You can just state the second harmonics frequency for the given wave in the question as

f_2 = \dfrac{1}{L} \sqrt{\dfrac{T}{\mu}}

or

\lambda_2 = \dfrac{1}{f_2} \sqrt{\dfrac{T}{\mu}}

where λ₂= L.
Identifying that the period of the 2 half-loop is the same, this implies that the frequency is the same, too.

I suspect that we can assume that the tension is the same in this situation because the mass per unit mass changes. If both tension and mass per unit mass change, I think the problem can be complicated to solve. We need to come back to solve the partial differential equation of the wave equation.

Assume that the tension is same, then the wavelength is proportional to

1/ \sqrt{\mu}

.
Since the wavelength increases from A to B as shown, the mass per unit length decreases from A to B so A has a greater diameter.

Reply 5

Yatayyat

OP

14

Original post by Eimmanuel

First of all, you should write the frequency of the normal modes of the standing wave is

f_n = \dfrac{n}{2L} \sqrt{\dfrac{T}{\mu}}

where n= 1, 2, 3…

if you want to write the general equation for the frequency instead of

You can just state the second harmonics frequency for the given wave in the question as

f_2 = \dfrac{1}{L} \sqrt{\dfrac{T}{\mu}}

or

\lambda_2 = \dfrac{1}{f_2} \sqrt{\dfrac{T}{\mu}}

where λ₂= L.
Identifying that the period of the 2 half-loop is the same, this implies that the frequency is the same, too.

I suspect that we can assume that the tension is the same in this situation because the mass per unit mass changes. If both tension and mass per unit mass change, I think the problem can be complicated to solve. We need to come back to solve the partial differential equation of the wave equation.

Assume that the tension is same, then the wavelength is proportional to

1/ \sqrt{\mu}

.
Since the wavelength increases from A to B as shown, the mass per unit length decreases from A to B so A has a greater diameter.

So just to reiterate it is true that wavelength is inversely proportional to the square root of mass per unit length (when constants are taken out).

And mass per unit definitely has to lower in order for wavelength to increase as shown in the figure according to the relationship. Silly of me why I couldn't see that again in the first place. Hence diameter also goes up.

And also don't know why I also expressed wavelength to be written as 2L if it explicitly says the second harmonic so L in this case

Thanks so much for the help :smile:

Second harmonic waves HELP!

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