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A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance by `2.0 cm`. the motion of bar is continuous and is repeated regularly `125` times per second. If klthe distance between adjacent wave crests is observed to be `15.6 cm` and the wave is moving along positive `x-`direction, and at `t=0` the element of the string at `x=0` is at mean position `y =0` and is moving downward, the equation of the wave is best described byA. `y=(1 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`B. `y=(2 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`C. `y=(1 cm) cos [(40.3 rad//m) x-(786 rad//s)t]`D. `y=(2 cm) cos[(40.3 rad//m) x-(786 rad//s)t]` |
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Answer» Correct Answer - a Amplitude of wave, `A=(2.0cm)/(2)=1 cm` frequency of wave, `f=125 Hz` Wavelength of wave, `lambda=15.6 cm=0.156 m` Let equation of wave be, `y=A sin (kx-omega t+phi)` where `k=2pi//lambda=40.3 rad//m and omega=2pif=786 rad//s` Using initial conditions, `y(0,0)=0=A sin phi` and `(dely)/(delt)=Aomega cos philt0` we get, `phi=0` So, the equation of wave is `y=(1 cm) sin[(40.3 rad//m)x-(786 rad//s)t]` |
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