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The total mass of the string is 3.1 g and the total length of the string is 0.91 m.

Show that the tension in the string when it is sounding the harmonic shown in Figure 7 is about 70 N.

Harmonic shown in figure 7 is 3rd harmonic, here speed is 145m/s and frequency is 330Hz.

Show that the tension in the string when it is sounding the harmonic shown in Figure 7 is about 70 N.

Harmonic shown in figure 7 is 3rd harmonic, here speed is 145m/s and frequency is 330Hz.

(c) First harmonic frequency = 110 Hz T = 4 × 110^2× 0.66^2× ( 3.1 x10^-8/ 0.91 ) = 71.8 N

Original post by Musc

Why to the power of -8 and not to the power of -3

Pretty sure they meant to say -3 as I used that and got the right answer

Original post by Akeel is wrong

(c) First harmonic frequency = 110 Hz T = 4 × 110^2× 0.66^2× ( 3.1 x10^-8/ 0.91 ) = 71.8 N

I know this is an old thread but if you used that, you're finding the tension for the first harmonic and not the third??

Original post by Qxi.xli

I know this is an old thread but if you used that, you're finding the tension for the first harmonic and not the third??

His calculations are wrong.I don’t think it’s possible to write 3rd harmonic in terms of first harmonic since you have the square root which would mean that as you go up in harmonics there isn’t a linear pattern/increase.Personally I would just do it the normal way🤣

Original post by The A.G

His calculations are wrong.I don’t think it’s possible to write 3rd harmonic in terms of first harmonic since you have the square root which would mean that as you go up in harmonics there isn’t a linear pattern/increase.Personally I would just do it the normal way🤣

Hang on sorry what's the normal way😭 the mark scheme says 71.8N too😭

Original post by Qxi.xli

Hang on sorry what's the normal way😭 the mark scheme says 71.8N too😭

Don’t worry I’ll tell you,you see how you have the equation for fundamental harmonic.On the equation the 2L represents the wavelength since when a stationary wave is vibrating at fundament harmonic the length of the string is only half the wave length.Now all you have to do is rearrange the equation then sub values

Original post by The A.G

Don’t worry I’ll tell you,you see how you have the equation for fundamental harmonic.On the equation the 2L represents the wavelength since when a stationary wave is vibrating at fundament harmonic the length of the string is only half the wave length.Now all you have to do is rearrange the equation then sub values

But that would give the wrong answer?

You have to sub In the fundamental freq (lambda/2) which is 110 in this case, but the question is asking for the tension when it's 1.5 lambda? That's where I'm a bit confused..

(edited 3 years ago)

Original post by Qxi.xli

But that would give the wrong answer?

You have to sub In the fundamental freq (lambda/2) which is 110 in this case, but the question is asking for the tension when it's 1.5 lambda? That's where I'm a bit confused..

You have to sub In the fundamental freq (lambda/2) which is 110 in this case, but the question is asking for the tension when it's 1.5 lambda? That's where I'm a bit confused..

Original post by The A.G

thank youu.. so 1 lamda is 91/150 m, but then if I sub that in I get a crazy number?

sorry this question is hella confusing for no reason

Original post by Qxi.xli

thank youu.. so 1 lamda is 91/150 m, but then if I sub that in I get a crazy number?

sorry this question is hella confusing for no reason

sorry this question is hella confusing for no reason

Could you show me what you did so I can check your numbers

Original post by The A.G

Could you show me what you did so I can check your numbers

sorry it's messy

Original post by Qxi.xli

sorry it's messy

I mean is still better than my handwriting 🤣.Anyways use the wave speed formula to find wavelength since it’s not stated that they used the whole of the 0.91 to form the Harmonic.

Original post by The A.G

I mean is still better than my handwriting 🤣.Anyways use the wave speed formula to find wavelength since it’s not stated that they used the whole of the 0.91 to form the Harmonic.

I looked up the question for you.It shows that only 0.66m was used to form the third harmonic

Original post by The A.G

I looked up the question for you.It shows that only 0.66m was used to form the third harmonic

that gives the right answer now, thanks!

Original post by username3477548

that gives the right answer now, thanks!

hey, i too have the same problem 2 years after. can you run me through your values and why you put them where/how they are? I understand the formula but my main issue is where 150 cam from and why you havent used 0.91m, but 91cm. thank you in advance

(edited 11 months ago)

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