Length and antinodes?

Question

Hi,

A steel wire is used to stretch a spring. An oscillating magnetic field drives the steel wire back and forth. A standing wave with three antinodes is created when the spring is stretched 7.5 cm. What stretch of the spring produces a standing wave with five antinodes?

I know that the length has to be shorter than 7.5 cm. How would I go about calculating it?


Thanks!

Answer 1

Hmmm...
If I'm correct the 3 antinodes at 7.5 cm apart provide you with a length of the spring L=4 x 7.5=30 cm

Then for five antinodes we have l=L/6=30/6=5cm

I'm just kidding

We have a constant frequency if oscillation f=const.
v=l f
l - wavelength (lambda)
v- speed of the wave

also we know that if L0 is the original length
Then
l1=2( L0+7.5 cm)/3 3 antinodes
l2=2(Lo+ 7.5 + x)/5
x - is what you have to find

we know that
v=sqrt( T/(m/L))
m - mass of the spring is the same
but tension T=-kd
and L or even original Lo remains unknown

we know that f1=f2
then

0.5 sqrt(T1/(m/L1)/ L1= 0.5 sqrt(T2/(m/L2)/ L2

T1/(L1) = T2/L2
k(y+7.5)/L1 = k(y+7.5 + x)/L2
(y+7.5)/L1 = (y+7.5 + x)/L2

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