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A travelling harmonic wave on a string is described by `y(x,t)=7.5sin(0.0050x +12t_pi//4)` (a) what are the displacement and velocity of oscillation of a point at `x=1cm, ` and `t=1s` ? I sthis velocity equal to the velocity of wave propagation ? (b) Locate the point of the string which ahve the same transverse displacement and velocity as `x=1cm` point at `t=2s, 5s and ` 11s. |
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Answer» The travelling harmonic wave is `y(x,t)=7.5sin (0.0050x +12t+pi//4)` At `x=1cm `and `t=1 sec, y(1,1)=7.5sin(0.005xx1+12xx1+pi//4)=7.5sin(12.005+pi//4)` …(i) Now, `theta=(12.005+pi//4) ` radian `=(180)/(pi)(12.005+pi//4)degree=(12.005xx180)/((22)/(7))=732.55^(@)` `:.` from (i), `y(1,1)=7.5 sin (732.55^(@))=7.5sin(720+12.55^(@))` `=7.5sin12.55^(@)=7.5xx0.2173=1.63cm` velocity of oscillation, `upsilon(d)/(dt)[y(1,1)}=(d)/(dt)[7.5sin(0.005x+12t+(pi)/(4))]` `=7.5xx12cos[0.005+12t+(pi)/(4)]` At, `x=1cm, t=1sec. ` `upsilon=7.5xx12cos(0.005+12+pi//4)=90cos(732.55^(@))=90cos(720+12.55^(@))` `upsilon=90cos(12.55^(@))=90xx0.9765=87.89cm//s`. Comparing the given eqn. with the standard form `y(x,t)=rsin[(2pi)/(lambda)(upsilont+x)+phi_(0)]` we get, `r=7.5, (2piupsilon)/(lambda)=2piv=12 or v=(6)/(pi)` `(2pi)/(lambda)=0.005. ` `:. lambda=(2pi)/(0.005)=(2xx3.14)/(0.005)=1256cm=12.56m` velocity of wave propagtion, `upsilon=vlambda=(6)/(pi)xx12.56=24m//s` We find that velocity at `x=1cm, 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 same transverse displacement and velocity . As `lambda=12.56m` therefore, all points at distance `+-12.6m, +-25.2m, +-37.8m, ...` from `x=1cm` will have same displacement and velocity, `x=1cm` point `t=2s`, 5s and 11s. |
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