Question

1. Transverse Mechanical Wave A transverse wave on a cord is given by D(x,t)=0.20 sin(3.0x-‐‐15.0t),where D and x are in m and t is in sec. (a) Calculate the period of the wave motion.(b) Calculate the propagation speed of the wave. (c) Calculate the acceleration of the particle on the cord at x=1 m and t=0.2 sec. Whatwould be its maximum, possible acceleration?(d) If the cord is under the constant tension of F=25.0 N, what is its linear massdensity?

Answer 1

Solution:

The equation of a transverse mechanical wave on the cord is given by,

D(x, t) = (0.20m)*sin(3.0*x – 15.0*t) -------------------------------------------------(1)

where D, x are in meter, and t is in seconds.

The general form of a transverse mechanical wave tavelling along X direction is given by,

y = y_max*sin(k*x – ω*t)                 -------------------------------------------------(2)

where ω = angular velocity in rad/s = 2π/T

k = angular wavenumber = 2π/λ

Although the wave moves in the X direction, the pointson the cord move in the vertical Y direction (that is perform up-down motion or transverse motion).

(a)

By comparing equation (1) and (2), we have,

ω = 15.0 rad/s

2π/T = 15.0 rad/s

T = 2*3.1416 rad/(15.0 rad/s)

T = 0.4188 s

Hence the period of wave motion is 0.42 s

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(b)

The propagation speed of the wave is given by,

v = ω/k

from equation (1) and (2) we have,

ω = 15.0 rad/s and k = 3.0 rad/m

v = (15.0 rad/s) / (3.0 rad/m)

v = 5.0 m/s

Hence the speed of the wave is 5.0 m/s

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(c) Although wave is moving in X direction, the points on the cord are oscillating in the Y direction.

Thus to find the acceleration of the particles on the cord we have to take two times partial derivative with respect to time of the equation (2) since,

y’ = ∂y/∂t is the velocity of the particles in Y direction and

y” = ∂y’/∂t is the acceleration of the particle in Y direction.

y’ = ∂[y_max*sin(k*x – ω*t)]/∂t

y’ = -ω* y_max*cos(k*x – ω*t)                 

(we have kept displacement x constant while taking derivative with respect to time t alone)

Further taking partial derivative one more time,

y” = ∂y’/∂t

y” = ∂[-ω* y_max*cos(k*x – ω*t)]/∂t

y” = -ω{ω*y_max*sin(k*x – ω*t)}

y” = -ω²*y_max*sin(k*x – ω*t)

Hence the acceleration y” at x = 1 m and t = 0.2 s is

y” = -(15.0rad/s)²*(0.20m)*sin(3.0*(1 m) – 15.0*(0.2 s))

y” = -(15.0rad/s)²*(0.20)*(0)

y” = 0.0m/s²

The maximum possible acceleration is achieved when sin(3.0*(1 m) – 15.0*(0.2 s) = 1

y_max” = -(15.0rad/s)²*(0.20)*(1)

y_max” = -45.0 m/s²

Hence the acceleration of the particle at x = 1m and t = 0.2 s is 0.0 m/s and the maximum possible acceleration is -45.0 m/s²

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(d)

The tension in the cord is F = 25.0 N

we have,

v = √[F/μ]                 where μ = linear mass density

v² = F/μ

μ = F/v²

μ = (25.0 N)/(5.0 m/s)²

μ = 1.0 kg/m

Hence the linear mass density is 1.0 kg/m