Saved Bookmarks
| 1. |
A sine wave has an amplitude A and wavelength `lambda`. Let V be the wave velocity and v be the maximum velocity of a particle in the medium. ThenA. V cannot be equal to vB. `V=v`, if `A=lambda/(2pi)`C. `V-v`, if `A=2 pi lambda`D. `V=v`, if `lambda =A/pi` |
|
Answer» Correct Answer - B Let the equation of wave by `y=A sin (omega t-kx)` where, `omega=2pi n` and `k=(2pi)/lambda`. Wave velocity, `V=n lambda =omega /(2pi)xx(2pi)/k=omega/k` Maximum particle velocity `v= A omega` For `V=v, omega/k=A omega` or `A=1/k=lambda/(2pi)`. |
|