InterviewSolution
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InterviewSolution
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Consider a standing wave formed on a string . It results due to the superposition of two waves travelling in opposite directions . The waves are travelling along the length of the string in the `x` - direction and displacements of elements on the string are along the `y` - direction . Individual equations of the two waves can be expressed as `Y_(1) = 6 (cm) sin [ 5 (rad//cm) x - 4 ( rad//s)t]` `Y_(2) = 6(cm) sin [ 5 (rad//cm)x + 4 (rad//s)t]` Here `x` and `y` are in `cm`. Answer the following questions. Amplitude of simple harmonic motion of a point on the string that is located at `x = 1.8 cm` will beA. `3.3 cm`B. `6.7 cm`C. `4.9 cm`D. `2.6 cm` |
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Answer» Correct Answer - C `y = y_(1) + y_(2) = ( 12 sin 5 x) cos 4 t` Maximum value of `y` - position in `SHM`of an element of the string that is located at an antinode `= +- 12 cm ( sin 5 x = +- 1)` For the position nodes amplitudeshould be zero. So , `sin 5 x = 0 rArr 5 x = n pi` `x = ( n pi)/(5)` where `n = 0 , 1 ,2 , 3 ,.....` Value of amplitude at `x = 1.8 cm` `A = 12 sin ( 5 xx 1.8) = 4.9 cm` At any instant say `t = 0` , instantaneous velocity of points on the string is zero for all points as at extreme position velocities of particles are zero. |
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