A travelling harmonic wave on a string is described by y(x, t)=7.5sin(0.0050x+12t+π4).
(a) what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 sec? 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 that of point at x = 1 cm and t = 2 s, 5 s and 11 s.
The given harmonic wave is:
y(x, t)=7.5 sin(0.0050x12t+π4)
For x = 1 cm and t = 1s,
y(1,1)=7.5sin(0.0050+12+π4)
=7.5sin(12.0050+π4)
=7.5sinθ
where, θ=12.0050+π4=12.0050+3.144=12.79 rad
=1803.14×12.79=732.81∘
∴ y(1,1)=7.5sin(732.81∘)=7.5sin(90×8+12.81∘)=7.5sin12.81∘=7.5×0.2217=1.6629≈1.663cm
The velocity of the oscillation:
v=ddty(x,t)=ddt[7.5sin(0.0050x+12t+π4)]=7.5×12cos(0.0050x+12t+π4)
At x=1 cm and t=1sec:
v(1, 1)=90cos(12.005+π4)=90cos(732.81∘)=90cos(90×8+12.81∘)=90cos(12.81∘)=90×0.975=87.75cm/s
Now, comparing the given equation with the equation of a propagating wave:
y(x, t)=a sin(kx+ωt+ϕ)
So,
ω=12 rad/s
k=0.0050 m-1
∴ v=120.0050=2400cm/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 given by:
k=2πλ
∴ λ=2πk=2×3.140.0050=1256 cm=12.56 m
All the points at distances nλ, i.e. ±12.56 m, ±25.12 m, ... where n=±1,±2,... and so on
will have the same displacement as that of he point at x=1 cm at t=2 s, 5 s, and 11 s.
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