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प्रश्न
The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is 360 ms–1 and their frequencies are 256 Hz.
- Calculate the time at which the second curve is plotted.
- Mark nodes and antinodes on the curve.
- Calculate the distance between A′ and C′.
उत्तर
Given the frequency of the wave f = 256 Hz
Time period T = `1/f`
= `1/256` s
= 3.9 × 10–3s
a. In second plot, the displacement of each particle is zero. It means all the points are crossing the mean position. At first plot point at A1 is at amplitude position. The time is taken to move from amplitude position to mean position is equal to one-fourth of the time period.
Hence, t = `T/4`
= `1/40`
= `(3.9 xx 10^-3)/4` s
= 9.8 × 10–4s
Method 2: Wavelength λ = `v/f`
= `(360 m/s)/(256 Hz)`
= 1.406 m
AA' = `λ/4`
= `(1.406 m)/4`
= 0.3516 m
Time (t), at which the second curve is plotted
t = `(A A^')/v`
= `(0.3516 m)/(360 m/s)`
= 0.000976 s
= 9.8 × 10–4s
b. Nodes are A, B, C, D, E (i.e., zero displacements)
Antinodes are A’, C’ (i.e., maximum displacement)
c. It is clear from the diagram A’ and C’ are consecutive antinodes, hence separation = wavelength
⇒ λ = `v/f = 360/256` = 1.41 m
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