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प्रश्न
The transverse displacement of a string (clamped at its both ends) is given by
y(x, t) = 0.06 sin `2/3` x cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10-2 kg.
Answer the following:
Interpret the wave as a superposition of two waves travelling in opposite
directions. What is the wavelength, frequency, and speed of each wave?
उत्तर १
The given equation is y(x, t) = 0.06 sin `(2π)/3 xx` cos 120 πt …(1)
We know that when a wave pulse
`"y"_1 = r sin (2pi)/lambda (vt - x)`
travelling along + direction of x-axis is superimposed by the reflected wave
`y_2 = -r sin (2pi)/lambda (vt + x)`
travelling in opposite direction,a stationary wave
`y = y_1 + y_2 = -2r sin (2pi)/lambda x cos (2pi)/lambda vt` is formed
Comparing equation (1) and (2) we find that
`(2pi)/lambda = (2pi)/3 => lambda = 3 m`
Also `(2pi)/lambda v = 120 pi` or `v = 60lambda = 60 xx 3 = 180 "ms"^(-1)`
Frequency, `v = v/lambda = 180/3` = 60 Hz
Note that both the waves have same wavelength same frequency and same speed.
उत्तर २
A wave travelling along the positive x-direction is given as:
`"y"_1 a sin(omega"t" - kx)`
The wave travelling along the negative x-direction is given as:
`y_2 = asin(omegat + kx)`
The superposition of these two waves yields:
`y = y_1 + y_2 = asin(omegat - kx) - asin(omegat + kx)`
`= asin(omegat)cos(kx) - asin(kx) cos(omegat) - asin(omegat)cos(kx) - asin(kx) cos(omegat)`
`= -2asin (kx)cos(omegat)`
`= -2asin((2pi)/lambda x) cos(2pivt)` ....(i)
The transverse displacement of the string is given as:
`y(x,t) = 0.06 sin((2pi)/3 x) cos(120pi t)` ....(ii)
Comparing equations (i) and (ii), we have:
`(2pi)/lambda = (2pi)/3`
∴Wavelength, λ = 3 m
It is given that:
120π = 2πν
Frequency, ν = 60 Hz
Wave speed, v = νλ
= 60 × 3
= 180 m/s
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संबंधित प्रश्न
The transverse displacement of a string (clamped at its both ends) is given by
y(x, t) = 0.06 sin `2/3` x cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is `3.0 xx 10^(-2)` kg.
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