Understanding Traverse Wave Properties: Homework Questions

  • Thread starter Thread starter GGD
  • Start date Start date
  • Tags Tags
    String Wave
AI Thread Summary
The discussion revolves around solving homework questions related to the properties of transverse waves on a rope. The amplitude of the motion was determined to be 0.199 m using the wave equation. Participants clarified that the wave speed should be calculated using the relationship between wavelength and frequency, leading to a corrected speed of 10 m/s. The tension in the rope can then be found using the wave speed and mass per unit length. The conversation highlights the importance of correctly applying wave equations and understanding fundamental mode characteristics.
GGD
Messages
3
Reaction score
0

Homework Statement



In a demonstration, a 0.45 kg horizontal rope is fi xed in place at its two ends (x = 0 and x = 5.0 m) and made to oscillate up and down in the fundamental mode, at frequency 4.0 Hz. At t = 0, the point at x = 2.5 m has zero displacement and is moving upward in the positive direction of a y-axis with a transverse velocity of 5.0 m/s.

(a) What is the amplitude of the motion of that point?
(b) What is the wave speed in the rope?
(c) What is the tension in the rope?


Homework Equations



w=2*pi*f
k=2*pi*λ
y(x,y)=Asin(wt-kx)
v=sqrt(T/μ)
v=λ*f

The Attempt at a Solution



I already found to amplitude to be .199 m by taking the derivite of the wave equation and solving for A.

For part B, though the wave speed would simply be wavelength*frequency, but 20m/s is not the correct answer. I tried to mess around with a few other equations but they led me nowhere.

And for part C I know T is simply (v^2)*μ, but I obviously need to first find velocity.

Any help would be grealty appreciated, thanks!
 
Physics news on Phys.org
What value did you use for the wavelength. The vibration is the fundamental which means the string is half a wavelength long .
 
Alright so the speed should equal ((wavelength*freq)/2), or (5*4)/2 , which is equal to 10 m/s? That answer doesn't work for me either.
 
First of all : you know the frequency is 4Hz and the wavelength is (2 x 5) =10m... this gives you the speed (v = fλ)
Speed = sqrt (T/μ) ...you should be able to get T ?
The amplitude is given by a SHM equation. Max velocity in SHM = ωA = 2∏fA
The midpoint of the string is doing SHM with max amplitude...
Hope this helps you to finish it.
 
Alrigh I finally got it, I was dividing by 2 instead of multiplying by it. Thanks for your help!
 
Great... well done
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top