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प्रश्न
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 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.
उत्तर १
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.
उत्तर २
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.
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