The equation of a wave travelling on a stretched string is y-4sin 2-t-0-02-x-100 here x- y are in cm and t in seconds- The relative deformation amplitude of medium is

A wave disturbance in a medium is described by y(x , t)= 0.02 cos (50 $π$ t + $\frac{π}{2}$ ) cos (10 $π$ x) where x and y are in metre and t in second

Q. The equation of a wave propagating along a stretched string is given by

Y = 4 sin 2 π [(t / 0.02) - (x / 100)]

, where

Y

and

x

are in cm and

t

in second. Determine velocity of wave.

Q. A wave disturbance in a medium is described by

y (x, t) = 0.02 cos (10 π x) cos (50 π t + \frac{π}{2})

, where

x

and

y

are in

meter

and

t

seconds

. Identify the incorrect option.

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The equation of a wave travelling on a stretched string is y=4sin2π(t0.02−x100) here x, y are in cm and t in seconds. The relative deformation amplitude of medium is

The correct option is A 0.08πy=4sin 2π(t0.02−x100)dydx=8π100 cos 2π(t0.02−x100)Deformation amplitude A=∣∣∣dydx∣∣∣=8π100=0.08π

The equation of a wave travelling on a stretched string is $y = 4 sin 2 π (\frac{t}{0.02} - \frac{x}{100})$ here x, y are in cm and t in seconds. The relative deformation amplitude of medium is

The correct option is A 0.08 $π$
$y = 4 s i n 2 π (\frac{t}{0.02} - \frac{x}{100})$
$\frac{d y}{d x} = \frac{8 π}{100} c o s 2 π (\frac{t}{0.02} - \frac{x}{100})$
Deformation amplitude $A = ∣ ∣ ∣ \frac{d y}{d x} ∣ ∣ ∣ = \frac{8 π}{100}$ $= 0.08 π$