Speed of longitudinal wave 30 times the speed of a transverse wave?by PirateFan308 Tags: longitudinal wave, transverse wave, waves, young's modulus 

#1
Mar2412, 10:33 AM

P: 94

1. The problem statement, all variables and given/known data
What must be the stress (F/A) in a stretched wire of a material whose Young's modulus is Y for the speed of longitudinal waves to equal 30 times the speed of transverse waves? 2. Relevant equations [itex]Y=\frac{Fl_0}{Al}[/itex] [itex]v_L=f\lambda = \sqrt{F/\mu}[/itex] [itex]v_T = \omega A sin(kx\omega t)[/itex] 3. The attempt at a solution I know that [itex]v_L=30v_T[/itex] but my main problem is that longitudinal velocity remains constant while transverse velocity is dependent on position and time, making it impossible for one to be a multiple of the other unless they are both equal to 0, which cannot be the case. I'm not sure what I'm missing ... thanks! 



#2
Mar2412, 12:15 PM

P: 383





#3
Mar2412, 12:21 PM

P: 383





#4
Mar2412, 01:20 PM

P: 94

Speed of longitudinal wave 30 times the speed of a transverse wave? 



#5
Mar2412, 01:25 PM

P: 297

Hint: You have missed a square root. 



#6
Mar2412, 01:31 PM

P: 94





#7
Mar2412, 02:24 PM

P: 94

So if [itex]v_L=\sqrt{Y/\rho}[/itex] and [itex]v_t=\sqrt{F/\mu}[/itex] then
[itex]\sqrt{Y/\rho}=30\sqrt{F/\mu}[/itex] which is equivalent to [itex]Y=\frac{900F\rho}{\mu}[/itex] [itex]Y=\frac{Fl_0}{Al}[/itex] so [itex]\frac{F}{A}=\frac{Yl}{l_0}=\frac{900Fl\rho}{l_0 \mu}[/itex] but [itex]l\rho = A[/itex] and [itex]\mu = \frac{mass}{length} ~~so~~\mu l_0=mass[/itex] so [itex]\frac{F}{A}=\frac{mass}{900AF} ~~so~~F^2=\frac{mass}{900}[/itex] which doesn't work ... I also tried it a different way, letting [itex]v_T=fλ[/itex] so that [itex]Y=900v^2\rho[/itex] so then [itex]\frac{F}{A}=\frac{Yl}{l_0} = \frac{900v^2\rho l}{l_0} = \frac{900v^2A^2}{l_0}[/itex] which also doesn't work. Any help would be appreciated! Thanks!! 



#8
Mar2412, 03:35 PM

P: 297

Skip the step where you have got in Y=FL/AL(0).
You have no way of relating L and L(0). Also Lp=A is not right. p=M/V sp Lp=LM/V. But here L is change in length. So you can't write V=AL as L is change in length.So this doesn't help. Just focus on your relation Y/p=900F/u. Divide it by A on both sides Y/Ap=900F/Au. So F/A=Yu/900Ap Is there any way you can find out what u/Ap is ? ( Hint:You have applied a relation between u and l in your last post.That might help) 



#9
Mar2512, 10:38 AM

P: 94

I got it  thank you!!



Register to reply 
Related Discussions  
The Speed of a transverse wave  Introductory Physics Homework  1  
Speed of Transverse Wave  Introductory Physics Homework  4  
Speed of a Transverse Wave  Introductory Physics Homework  4  
Transverse wave speed  Introductory Physics Homework  3 