Two wires W1 and W2 have the same
radius r and respective densities ρ
1 and ρ2 such that ρ2=4ρ1. 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 W1 to W2 is :
(1) 1: 1
(2) 1: 2
(3) 1 : 3
(4) 4: 1
Asked by modi72879
2nd April 2018,
7:21 PM
Answered by Expert
Answer:
density of W2 is 4 times that of W1. Speed of waves in a stretched string inversly proportional to √μ, where μ is mass per unit length.
Hence Speed of waves in W1 and W2 is in the ratio 2:1, In W1 wave will move with high speed.
Since frequency of vibration is constant, to adjust for less speed in W2, wavelength has to be shorter( v = nλ, v is spped, n -frequency, λ is wavelength)
v1/λ1 = v2/λ2 ; v1/v2 = λ1/λ2 = 2;
hence ratio of number of antinodes in W1 and W2 is 1:2
Answered by Expert
3rd April 2018,
11:35 AM
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