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 :

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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