Waves

JEE Main level, single correct, medium

60 of 142

Question 60 of 142

Two wires W1{W}_{1} and W2{W}_{2} have the same radius rrand respective, densities ρ1{\rho }_{1} and ρ2{\rho }_{2}, such that ρ2=4ρ1{\rho }_{2}=4{\rho }_{1} . They are joined together at the point OO, as shown in the figure. The combination is used as a sonometer wire and kept under tension TT. The point OO 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 W1{W}_{1} to W2{W}_{2} is

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Answer: (B) 1:21:2

Let p1{p}_{1} and p2{p}_{2} be the number of antinodes of both wires.
Fundamental frequency remains the same for both wires.


n1=n2{n}_{1}={n}_{2}

T→sameT\rightarrow same

r→samer\rightarrow same

l→samel\rightarrow same

n=p2lTπr2ρn=\dfrac{p}{2l}\sqrt{\dfrac{T}{\pi {r}^{2}\rho }}
From the above equation, we have

p1ρ1=p2ρ2\dfrac{{p}_{1}}{\sqrt{{\rho }_{1}}}=\dfrac{{p}_{2}}{\sqrt{{\rho }_{2}}}
→p1p2=ρ1ρ2\rightarrow \dfrac{{p}_{1}}{{p}_{2}}=\sqrt{\dfrac{{\rho }_{1}}{{\rho }_{2}}}

The ratio of antinodes is
p1p2=12\dfrac{{p}_{1}}{{p}_{2}}=\dfrac{1}{2}.

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