Question 160: Waves

Question

Two wires W
1
and W
2
have the same radius r and respective densities $\rho $
1
and $\rho $
2
such that ρ
2
= 4$\rho $
1
. They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W
1
to W
2
is :
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l83nzr9f/2aa489a4-5800-4f4b-8b92-76f6d2c12b33/87497130-354b-11ed-8c5c-f3f0850e6b94/file-1l83nzr9g.png?format=png)

Accepted Answer

Fundamental frequency of a stretched string is ,
f = ${n \over {2L}}\sqrt {{T \over \mu }} $
Here,  n = number of antinodes
$\mu $ = mass per unit length.
As, O is the midpoint of two bridegs,  hence length of two wires are equal.
$	herefore\,\,\,$  L
1
= L
2
= L
As frequency of both wires same
f
1
= f
2
$ \Rightarrow $$\,\,\,$ ${{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{ho _1}}}} $ = ${{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{ho _2}}}} $
$ \Rightarrow $$\,\,\,$ ${{{n_1}} \over {{n_2}}}$ = $\sqrt {{{{ho _1}} \over {{ho _2}}}} $
$ \Rightarrow $$\,\,\,$ ${{{n_1}} \over {{n_2}}}$ = $\sqrt {{{{ho _1}} \over {4{ho _1}}}} $ = ${1 \over 2}$

Answer

1 : 1

Answer

1 : 2

Answer

1 : 3

Answer

4 : 1

Waves

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