String resonance with increased tension and standing wave pattern

  • Thread starter Thread starter Alt+F4
  • Start date Start date
  • Tags Tags
    Mass String
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 7K views
Alt+F4
Messages
305
Reaction score
0
A string is stretched between two supports that are L = 1.2 m apart. It resonates at a frequency of f = 450 Hz with a standing wave pattern that has two nodes between the two supports.



--------------------------------------------------------------------------------
(a)Using the unit of m , find the wavelength l.
Wavelength = .8
I did 1.2*2 /3 which is the correct answer then

(b) Suppose that the tension, T, in the string is increased by a factor of 4. What is the new frequency, f', in unit of Hz , if the string vibrates with the same standing wave pattern that it began with (i.e. A standing wave pattern that has two nodes between the supports).

HELP: The wave length l is determined by the number of nodes, so it keeps unchanged; however, the wave speed is determined by the tension T and the string's density d = mstring/L, so it will change.

HELP: First, find the incresing factor of the wave speed v by using the equation of v2 = T/d, and then find the incresing factor of the frequency by using the equation of v = f l. Remember, l = l' because the number of nodes is same.

I don't know what the mass of the string is so how do i find it inorder to get desity? Thanks
 
Physics news on Phys.org
Also this Problem
As you move toward one of the loudspeakers, the intensity decreases. When you have moved by 0.5 m , you hit the first point of zero intensity.

Assuming that the speed of sound is v = 330 m/s , what is the frequency f of the sound in Hz ?

okay The distance between two closest nodes in a standing wave pattern is lambda/2, where lambda is the wavelength. Hence, the distance from an intensity maximum to the closest node is lambda/4, which is the distance by which you have moved. This gives you lambda.

So what i did was .5/4 = .125
v=lambda*f.
330 = .125 f
Which is not the asnwer
 
The mass density of the string is the same (im guessing they don't want you to worry about the fact that the string will stretched lowering it's mass density). Think about how increasing the tension will SCALE the previous velocity and use that to find the new frequency.

for your second post, it looks like you made a typo. I think you meant to multiply by, not divide by four, to get the wavelength.
 
Last edited: