A wave pulse passing on a string with a speed of $40 c m s^{- 1}$ in the negative x-direction has its maximum at $x = 0$ at $t = 0$ . Where will this maximum be located at $t = 5 s$ ?

-100

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-200

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2000

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Solution

The correct option is B -200
speed of the wave is $(v) = 40 c m / s$
in time $t = 55$ ,the maximum will travel a distance
$d = v t$
$= 40 \times 5$
$= 200 c m / s$
now the wave travels at $(- x)$ direction.
so the maximum will beat :- $x = - 200 c m$

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A wave pulse passing on a string with a speed of 40 cm s−1 in the negative x-direction has its maximum at x=0 at t=0. Where will this maximum be located at t=5 s?

The correct option is B -200speed of the wave is (v)=40cm/s in time t=55,the maximum will travel a distanced=vt =40×5 =200cm/snow the wave travels at (−x) direction. so the maximum will beat :- x=−200cm

A wave pulse passing on a string with a speed of $40 c m s^{- 1}$ in the negative x-direction has its maximum at $x = 0$ at $t = 0$ . Where will this maximum be located at $t = 5 s$ ?

The correct option is B -200
speed of the wave is $(v) = 40 c m / s$
in time $t = 55$ ,the maximum will travel a distance
$d = v t$
$= 40 \times 5$
$= 200 c m / s$
now the wave travels at $(- x)$ direction.
so the maximum will beat :- $x = - 200 c m$