Class 11 Physics - Chapter Waves NCERT Solutions | The transverse displacement of a wire (c
The transverse displacement of a wire (clamped at both its ends) is described as :
y (x, t) = 0.06 sin(2π/3x) cos(120πt)
The mass of the wire is 6 x 10-2 kg and its length is 3m. Provide answers to the following questions:
( I ) Is the function describing a stationary wave or a travelling wave?
( ii ) Interpret the wave as a superposition of two waves travelling in opposite directions. Find the speed, wavelength and frequency of each wave.
( iii ) Calculate the wire’s tension. [X and y are in meters and t in secs]
We know,
The standard equation of a stationary wave is described as:
y (x, t) = 2a sin kx cos ωt
Our given equation y (x, t) =0.06sin(2π/3x)cos(120πt) is similar to the general equation .
( i ) Thus, the given function describes a stationary wave.
( ii ) We know, a wave travelling in the positive x – direction can be represented as :
y1 = a sin( ωt – kx )
Also,
Wave travelling in the negative x – direction is represented as :
y2 = a sin( ωt + kx )
Super- positioning these two waves gives us :
y = y1 + y2
= a sin( ωt – kx ) – a sin( ωt + kx )
= asin(ωt)cos(kx) – asin(kx)cos(ωt) – asin(ωt)cos(kx) – asin(kx)cos(ωt)
= – 2asin(kx)cos(ωt)
= −2asin(2πλx)cos(2πvt) . . . . . . . . . . . . . . ( 1 )
The transverse displacement of the wires is described as :
0.06sin(2π3x)cos(120πt) . . . . . . . . . . . . . . . . .( 2 )
Comparing equations ( 1 ) and ( 2 ) , we get :
2π/ λ = 2π/ 3
Therefore, wavelength λ = 3m
Also, 2πv/λ = 120π
Therefore, speed v = 180 m/s
And, Frequency = v/λ = 180/3
= 60 Hz
( iii )Given, Velocity of the transverse wave, v = 180 m / s
The string’s mass, m = 6 × 10-2 kg
String length, l = 3 m
Mass per unit length of the string, μ = m/l = (6 x 10-2 )/3
= 2 x 10-2 kg/m
Let the tension in the wire be T
Therefore, T = v2 μ
= 1802 x 2 x 10-2
= 648 N.
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