Class 11 Physics - Chapter Waves NCERT Solutions | A travelling harmonic wave on a string i
A travelling harmonic wave on a string is described by
y(x,t) = 7.5sin [0.0050x + 12t + π/4]
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
(a) The given harmonic wave is:
y(x,t) = 7.5sin [0.0050x + 12t + π/4]
For x = 1 cm and t = 1s,
y = (1, 1) = 7.5sin [0.0050 + 12 + π/4]
.= 7.5sin [12.0050 + π/4]
= 7.5 sinθ
Where, θ = 12.0050 + π/4 = 12.0050 + 3.14 / 4 = 12.79 rad
= 180 /3.14 x 12.79 = 732.81°
∴ y = (1, 1) = 7.5sin [732.81°]
= 7.5 sin (90 x 8 + 12.81°)
= 7.5 sin (12.81°)
= 7.5 x 0.2217
= 1.6629 ≈ 1.663 cm
The velocity of the oscillation at a given point and time is given as:
v = d/dt y(x,t) = d/dt [7.5sin(0.0050x + 12t +π/4)]
= 7.5 x 12cos (0.0050x + 12t +π/4)
At x = 1 cm and t = 1s:
v = y(1,1) = 90 cos (12.005 + π/4)
= 90cos(732.81°) = 90cos(90 x 8 + 12.81°)
= 90cos(12.81°)
= 90 x 0.975 = 87.75 cm/s
Now, the equation of a propagating wave is given by:
y(x,t) = a sin(kx + wt + ø)
Where,
k = 2π / λ
∴ λ = 2π / k
And ω = 2πv
∴ v = ω / 2π
Speed = v = vλ = ω / k
Where
ω = 12 rad/s
k = 0.0050 m-1
∴ v = 12 /0.0050 = 2400 cm/s
∴ Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.
(b) Propagation constant is related to wavelength as:
k = 2π / λ
∴ λ = 2π / k = 2 x 3.14 / 0.0050
= 1256 cm = 12.56
Therefore, all the points at distances nλ , (n =±1, ±2....and so on) i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.
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