Question

Consider the standing wave pattern with 4 antinodes. This standing wave is on a wire of linear mass density 3g/m and length 2m. It is being driven by a magnetic vibrator on the end of the wire, wiggling the wire up and down at 120 Hz, find tension. The answer must be in Newton. The tension is now slowly increased, causing the standing wave pattern shown to disappear. At certain higher tension the next standing wave pattern appears. Tell how many antinodes there are in the new standing wave and find new tension antinodes?

Answer 1

Part A

Velocity of waves on a string $V=\sqrt{\frac{T}{\mu}}$

There are Four antinodes.

Length of string is $L=2\,m$

Linear mass density $\mu=3\times10^{-3}\,kg/m$

since there are four antinodes, length of string is $L=4\times\frac{\lambda}{2}$

Hence wave length is $\lambda=\frac{L}{2}=1\,m$

Frequency is $f=120\,Hz$ , Hence wave speed is $V=\lambda{f}=1\times120=120\,m/s$

since $V=\sqrt{\frac{T}{\mu}}$ , $T=V^{2}\mu=(120)^{2}\times3\times10^{-3}=43.2\,N$

Part B

If there are antinodes on the string, $L=n\frac{\lambda}{2}=n\frac{V}{2f}$

$f=n\frac{V}{2L}$

Using the formula $V=\sqrt{\frac{T}{\mu}}$

$f=\frac{n}{2L}=\frac{n}{2L}\sqrt{\frac{T}{\mu}}$

Frequency, linear mass density and Length are not changed. That is $n\sqrt{T}$ does not change.

$n\sqrt{T}=n'\sqrt{T'}$

$4\sqrt{43.2}=n'\sqrt{T'}$

since Tension has increased That is .

Since the next lower integral number is 3, $n'=3$ .

New number of antinodes is 3.

$4\sqrt{43.2}=3\sqrt{T'}$

$T'=43.2\times\frac{4^2}{3^2}=76.8\,N$