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.

Question

Philoid · Accepted Answer

(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.