Vibration problem given Force,length and mass

[Equilibrium]

Homework Statement

A flexible wire 80 cm long has a mass of 0.40g. It is stretched across stops on a sonometer that are 50cm apart by a force of 500N. The wire may vibrate at the following frequencies except one. WHich one?
a. 4000vib/s
b. 1000 vib/s
c. 3000 vib / s
d. 1500 vib/s

Homework Equations

searched google for formula but none so far

no formula in my notes.
probably this one

$a = -\frac{k}{m}s$

s = displacement
m = mass
k = spring constant
a = acceleration

or this one
$T=2\pi\sqrt{\frac{m}{k}}$

T = period
m = mass
k = constant

The Attempt at a Solution

i don't know
can you help me start?

 

[BruceW]

You've written down the equations for simple harmonic motion. But I don't think that is what the question is about. If you look up a sonometer on google, it looks very much like the situation of standing waves on a string. (since it is fixed at two ends). So I think you should be using the equations for standing waves on a string.

 

[Equilibrium]

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.htmli see now but the answer is not given in the multiple choice
$$f=\frac{v}{\lambda}$$
$$f=\frac{v}{2L}\mbox{ since }L=\frac{\lambda}{2}$$
$$v=\sqrt{\frac{T}{\rho}}=\sqrt{\frac{T}{\frac{m}{L}}}$$
$$f=\frac{1}{2L}\sqrt{\frac{T}{\frac{m}{L}}}$$
$$f=625Hz=625vib/s$$
the question might be a typo or is it not?

 

[BruceW]

I think you might be using incorrect values for L or m. The entire wire is 80cm, with mass 0.4g. But only 50cm of that wire is within the sonometer. So what should you use for L and m in the equation?

 

[Equilibrium]

i see, so $$\rho=\frac{m}{L_{string}}$$ L_string = 80cm is different from
the length of the string apart which is the wavelength/2 = 50cm

1000vib/s
thanks

 

[BruceW]

yep. no worries :)