Question

A string of length L = 1.2 m is tied to a sinusoidal oscillator at one end, and is stretched by
a mass m hanging over a pulley on the other end. The frequency f of the oscillator is 120
Hz. Standing wave patterns appear when the mass of the hanging mass is 286.1 g and
447.0 g, but not for any intermediate mass. If the two ends can be taken as the nodes of the
standing waves, what is the linear density of the string?

Answer 1

what I want to ask is, is the tension exactly equal to mg?
I know \[f _{n}=\frac{ nv }{ 2L }=\frac{ n }{ 2L }\sqrt{\frac{ \tau }{ \mu }}\]
but \[\tau = mg ???\]

Answer 2

if not, how to find tension? @AllTehMaffs

Answer 3

I think since tension is the only force, and they don't give you a spring constant or anything for the string, then the Tension has to be mg like you said - I'm checking through some books real fast though...

Answer 4

I think you're right ^^

Answer 5

I get confused now.
how to make use of the 2 masses?

Answer 6

I'm not sure yet - it has something to do with figuring out which mode the string vibrates at (n) ....

Answer 7

I'm having trouble figuring out the integers that the two modes have to be - we know they have to be of the ratio
\[ \frac{n_1}{n_2} = \sqrt{\frac{T_2}{T_1}}\]
but nothing useful is coming out of it :P

Answer 8

:P 2861 is a prime number! blah

Answer 9

Can you still use that equation even though this is a forced oscillation?

Answer 10

~forced vibration

Answer 11

Yep I think so... Must still be valid but not sure if useful here
cannot manage to use 2 masses to find the mu

Answer 12

I have a super silly fix that's probably wrong!


\[ \frac{n_1}{n_2} = \sqrt{\frac{4470}{2861}}\]
therefor
\[ \frac{n_1^2}{n_2^2} = \frac{4470}{2861}\]
\[n_1^2 = 4470 \quad ; \quad n_2^2 = 2861\]
Then the equation
\[f=\frac{n}{2L} \sqrt{\frac{T}{\mu}}\]
\[ f^2 = \frac{n^2}{4L^2} \frac{T}{\mu}\]
\[ \mu =\frac{n_1^2}{4L^2} \frac{T_1}{f^2} \]
\[ = \frac{(4470)}{4(1.2m)^2}\frac{(286.1g)(9.81m/s^2)}{(120Hz)^2}\]

Answer 13

completely defeats the purpose of an integer mode :P

Answer 14

I dunno how else to use the masses though :/

Answer 15

what do ya think? ^_^

Answer 16

not quite agree with you though ;(
last part if you substitute n_1 and n_2 you will get different results @@
mu should be the same for both modes right?

Answer 17

I'm gettin' the same answer for each - 151.2 g/m

n1 = 4470 goes with m1 = 286.1
n2 = 2861 goes with m2 = 447

Answer 18

Still not at all sure that it's right, but they do agree with each other (forcibly so)

Answer 19

well the n ratio part is a bit weird
but at least they are forcibly agreeing as you said :P
I am thinking if it is physically correct

Answer 20

it seems like a pretty heavy string, but not outrageous

Answer 21

Hmm... it's about 3 times more dense than a really thick piano string

Answer 22

https://sites.google.com/site/lorenkoehlerwebsite/piano-craft/piano-wire

Answer 23

By the way it was mentioned
Take g=9.81ms-2
so I was wondering if it has to be used

Answer 24

it was in there ^_^
\[ = \frac{(4470)}{4(1.2m)^2}\frac{(286.1g)\overset{here}{(9.81m/s^2)}}{(120Hz)^2}\]

Answer 25

ahha sorry -v-

Answer 26

I will use your answer in the homework
Thank you for your help
hope it would be correct :)
(though I am not very sure even now :P

Answer 27

I hope so too, but I agree that it's a weird answer :P Hope I didn't hurt more than I helped!

Answer 28

just put an asterix saying that some crazy dude from online kept insisting that it was right ^_^

Answer 29

I remember one point which I think it is weird
the equation with f
how to ensure that the frequency is 120?
\[f _{n}=nv/2L \]

Answer 30

f may not be fundamental if n is a larger number
and uh I still find the n so wierd haha

Answer 31

Yeah, that's what I thought was weird too - I thought that the first standing waves wold be at n=1 and then just be some multiple of that for the larger mass. It's just not workin' out that way though :P Maybe because the frequency is being forced the harmonics are all messed up?

I don't know :/ this probably isn't right :( because doesn't n have to be an integer by definition?

Answer 32

my book says that n can go from 1 to infinity tho...

Answer 33

Since the problem says that both ends can be nodes though, the only other equation I have in my book for a forced vibration just turns into the one you posted. And the oscillator is what is ensuring f=120hz

Answer 34

I have had trouble with this, also. The velocity of the wave is the square root of the ratio of the tension to the mass per unit length of the string. The velocity is also the product of the frequency times the wavelength. The tension should be just the product of mg, mass times gravitational acceleration. If these are two nodes, then the longest wavelength will be a half wave over the 1.2m length, or 2.4m, and the next would be a full wavelength over 1.2m. That gives velocities in the ratio of 2:1 but we have tensions in the ratio 1.56:1, and their square roots in the ratio of 1.25:1 (=5/4). That suggests we are dealing with velocities in that ratio and thus wavelengths in that ratio inverse (4/5).
Perhaps "nodes" is used loosely and the masses are moving enough to be "antinodes"? Perhaps we are well away from the fundamental frequency? Must say 'goodnight."