JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 16)
A standing wave is formed by the superposition of two waves travelling in
opposite directions. The transverse displacement is given by
y(x, t) = 0.5 sin $$\left( {{{5\pi } \over 4}x} \right)\,$$ cos(200 $$\pi $$t).
What is the speed of the travelling wave moving in the positive x direction ?
(x and t are in meter and second, respectively.)
y(x, t) = 0.5 sin $$\left( {{{5\pi } \over 4}x} \right)\,$$ cos(200 $$\pi $$t).
What is the speed of the travelling wave moving in the positive x direction ?
(x and t are in meter and second, respectively.)
160 m/s
90 m/s
180 m/s
120 m/s
설명
Standard equation of standing wave,
y(x, t) = 2a sin kx cos $$\omega $$t
Given,
y(x, t) = 0.5 sin $$\left( {{{5\pi } \over 4}x} \right)$$ cos (200$$\pi $$t).
So, k = $${{5\pi } \over 4}$$ and $$\omega $$ = 200$$\pi $$
$$\therefore\,\,\,$$ Speed of travelling wave
= $${\omega \over k}$$ = $${{200\pi } \over {{{5\pi } \over 4}}}$$ = 160 m/s.
y(x, t) = 2a sin kx cos $$\omega $$t
Given,
y(x, t) = 0.5 sin $$\left( {{{5\pi } \over 4}x} \right)$$ cos (200$$\pi $$t).
So, k = $${{5\pi } \over 4}$$ and $$\omega $$ = 200$$\pi $$
$$\therefore\,\,\,$$ Speed of travelling wave
= $${\omega \over k}$$ = $${{200\pi } \over {{{5\pi } \over 4}}}$$ = 160 m/s.