Question

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.

Answer 1

a)

The given wave equation is,

y( x,t )=7.5sin( 0.005x+12t+ π 4 )

By substituting x=1 and t=1 in the above expression, we get

y=7.5sin( 0.005×1+12×1+ π 4 ) =7.5sin( 12.79 rad ) =7.5sin( 732.81° ) =1.663 cm

The velocity of oscillations at a given point is given as,

v= dy dt = d dt ( 7.5sin( 0.005x+12t+ π 4 ) ) =7.5×12cos( 0.005x+12t+ π 4 )

By substituting x=1 and t=1 in the above expression, we get

v=7.5×12cos( 0.005×1+12×1+ π 4 ) =90cos( 12.79 rad ) =90cos( 732.81° ) =87.76 cm/s

The equation of propagating wave is given by,

y( x,t )=asin( kx+ωt+ϕ )

Where,

k= 2π λ λ= 2π k

Also,

ω=2πν ν= ω 2π

Speed is given by,

Speed=λν = ω k

Here, ω=12 rad/s and k=0.0050  cm −1

By substituting the values of ω and k in the speed equation, we get

v= 12 0.0050 =2400 cm/sec

So, the velocity of wave propagation and wave oscillation is not equal.

Thus, the displacement of oscillation of appoint is 1.663 cm and the velocity is 87.76 cm/s.

b)

The wavelength of oscillation is given as,

λ= 2π k

By substituting the values in the above equation, we get

λ= 2×3.14 0.0050 =1256 cm =12.56 m

Thus, all the points at distance nλ that is ±12.56 m,±25.12 m,..... and so on for x=1 will have the same displacement as the x=1 cm points at t=2 s,5 s and 11 s.