Standing waves question

1. Nov 23, 2006

limeater

A string of fixed length L=1.200m is vibrated at a fixed frequency of f=120.0Hz. The tension, Ts, of the string can be varied. Standing waves with fewer than seven nodes are observed on the string when the tension is 2.654N and 4.147N, but not for any intermediate tension. What is the linear density of the string?

Hey. this I've been trying to understand this question for over half an hour now with no luck. what does it mean with "fewer than 7 nodes".
Any help is greatly appreciated

2. Nov 23, 2006

Kurdt

Staff Emeritus
3. Nov 23, 2006

limeater

yeah i understand what nodes are and also looked through the link. I also understand that a standing wave can exist on a string only if its wavelength can be given by the equation:

wavelength= 2 x length of string / m
m being an integer >0 and also the mode of the string

the mode in this case can be anything between 1 and 8
im still lost though

4. Nov 23, 2006

Kurdt

Staff Emeritus
Well from the link it gives you the equation for the natural frequency of a string including the mass which is unknown. everything else is apart from the mode of oscillation. Its easy to modify the equation to dea with other modes. Then you just play around and see which two modes with the two tensions give you the same value for mass.

Also remember linear density is m/L.

5. Nov 23, 2006

limeater

ok so would this be a correct way of approaching it?

by combining f=mv/2L and v=srqt(T/u)
m being the mode, T being the tension and u being the linear density
we can derive u=(T)(m^2)/(2Lf)^2
with the values given we have
u=(T)(m^2)/82944

Now I make a table

------T=2.654N-------T=4.147N
m=1
m=2
.
.
.
m=8

Try and match two values of u on the two columns and I will have my answer?

6. Nov 23, 2006

Kurdt

Staff Emeritus
Yeah you can do it that way.

7. Nov 23, 2006

limeater

i thought of another way of doing it which relates to the chart

consider x and y to be 2 different modes between 1 and 8

then
(2.654)(x^2)=(4.147)(y^2)
x/y=srt(4.147/2.654)
x/y=1.25
x/y=5/4
meaning that the linear density is equal at mode 5 and tension 2.654 to the linear density at mode 4 and tension 4.147
linear density=8.0x10^-4

The problem is that x/y is not exactly equal to 1.25
its more like 1.25002, so will my method be still correct?

8. Nov 24, 2006

Kurdt

Staff Emeritus
Thats how I did it.

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