A travelling harmonic wave on a string is described by

y(x, t) = 7.5 sin (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=1cm and t=1s,


y(1,1) = 7.5sin(0.005×1 + 12×1 + π/4)


y(1,1) = 7.5sin(12.005 + π/4)


y(1,1) = 7.5sin(12.79)


y(1,1) = 7.5 × 0.2217


y(1,1) = 1.663 cm


The velocity of oscillation is calculated as






At x=1cm and t=1s,


v(1,1) = 90cos(0.005×1 + 12×1 + π/4)


v(1,1) = 90cos(12.005 + π/4)


v(1,1) = 7.5cos(12.79)


v(1,1) = 7.5 × 0.975


v(1,1) = 87.75 cm/s


The equation of a propagating wave is given by


y(x,t) = asin(kx + wt + Φ)


where, k = 2π/λ


λ = 2π/k


and ω = 2πf


f = ω/2π


Speed, v = fλ = ω/k


where, ω = 12rad/s k = 0.0050 m-1


v = 12/0.0050


v = 2400 cm/s


Hence, the velocity of the wave oscillation at x = 1 cm and t = 1s is not equal to the velocity of the wave propagation.


(b) Propagation constant is related to wavelength as: k = 2π/λ


λ = 2π/k


λ = 2 × 3.14/0.0050


λ = 1256 cm


λ = 12.56 m


Therefore, all the points at distance nλ (n = 1, 2, .....), 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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