# A Travelling Harmonic Wave on a String is Described by What Are the Displacement and Velocity of Oscillation of a Point at X = 1 Cm, and T = 1 S? - Physics

A travelling harmonic wave on a string is described by

y(x,t) = 7.5 sin (0.0050x + 12t + pi/4)

(a) What are the displacement and velocity of oscillation of a point at = 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.

#### Solution 1

a) The given harmonic wave is

y(x,t) = 7.5 sin (0.0050x + 12t + pi/4)

For x = 1 cm and t = 1 s

y = (1,1) = 7.5 sin (0.0050 + 12 + pi/4)

= 7.5 sin(12.0050 + pi/4)

= 7.5 sin theta

Where, theta = 12.0050 + pi/4 = 12.0050 + 3.14/4 = 12.79

= 180/3.14 xx 12.79 = 732.81^@

:. y = (1,1) = 7.5 sin (732.81^@)

= 7.5 sin (90xx 8 + 12.81^@) = 7.5 sin 12.81^@

= 7.5 xx 0.2217

= 1.6629 ~~ 1.663 cm

The velocity of the oscillation at a given point and time is given as:

v = d/dt y(x,t) = d/(dt) [7.5 sin(0.0050x + 12t + pi/4)]

= 7.5 xx 12 cos (0.0050x + 12t + pi/4)

At x = 1 cm and t = 1s

v = y(1,1) = 90 cos (12.005 + pi/4)

= 90 cos(732.81^@) = 90 cos (90 xx 8 + 12.81^@)

=90cos(12.81^@)

=90xx0.975 = 87.75 cm/s

Now, the equation of a propagating wave is given by:

y(x,t) = a sin(kx + wt + phi)

where

k = (2pi)/lambda

:. lambda = (2pi)/k

And omega = 2pi/k

:. v = omega/(2pi)

Speed,  = vlambda = omega/k

where

omega = 12 "rad/s"

k = 0.0050 m^(-1)

:. v = 12/0.0050 = 2400 "cm/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 related to wavelength as:

k = (2pi)/lambda

:. lambda = 2pi/k = (2xx 3.14)/0.0050

= 1256 cm = 12.56 m

Therefore, all the points at distances 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 att = 2 s, 5 s, and 11 s.

#### Solution 2

THe travelling harmonic wave is y (x,t) = 7.5 sin(0.0050x + 12t + pi/4)

At x = 1 cm and t =1 sec

y(1,1)= 7.5 sin(0.005 xx 1 + 12 xx 1 pi/4) = 7.5 sin (12.005 + pi/4)  ...(i)

Now 'theta = (12.005 + pi/4) radian"

=  180/pi (12.005 = pi/4) "degree" = (12.005xx180)/(22/7)+ 45 = 732.55^@

:. From (i), y(1,1) = 7.5 sin(732.55^@) = 7.5 sin(720 + 12.55^@)

= 7.5  sin 12.55^@ = 7.5 xx 0.2173 = 1.63 cm

"Velocity of oscillation", v = (dy)/(dt) (1,1) = d/dt[7.5 sin(0.005x + 12 + pi/4)]

= 7.5 xx 12 cos [0.005x + 12t + pi/4]

At x = 1 cm, t = 1 sec

v = 7.5 xx 12 cos (0.005 + 12 + pi/4) = 90 cos(732.35^@)

= 90 cos (720 + 12.55)

v = 90 cos(12.55^@) = 90 xx 0.9765 = 87.89  cm/s

Comparing the given equation with the standard form y(x,t) = t sin[pi/4 (vt+x) + phi_0]

We get r = 7.5 , (2piv)/lambda = 12  or 2piv = 12

v = 6/pi

(2pi)/lambda = 0.005

:. lambda = (2pi)/0.005 = (2xx3.14)/(0.005) = 1256 cm = 12.56 m

Velocity of wave propagation, v = vlambda = 6/pi xx 12.56 = 24` m/s

We find that velocity at x =1 cm, t = 1 sec is not equal to velocity of wave propagation

b) Now all points which are at a distance of +- lambda +-2lambda, +-3lambda from x =1 cm will have sametransverse displacement and velocity. As lambda = 12.56 m, therefoe all  points at distances +- 12.6 m +- 25.2 m , +- 37.8 m .. form x = 1 cm will have same displacement and velocity as at x = 1 point t = 2 s, 5 s and 11 s.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 15 Waves
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