A wave travelling on a string stretched along the X -axis has an equation given by y = A\, e^ -((x)/(a) + (t)/(T))^2 . Identify the direction in which the wave is travelling.

Correct answer: B. Negative x -direction The given equation of the wave is y = A e^ -((x)/(a) + (t)/(T))^2 . For a point on the wave with a constant phase, we have: (x)/(a) + (t)/(T) = constant Differentiating with respect to time t , we get: (1)/(a) (dx)/(dt) + (1)/(T) = 0 (dx)/(dt) = -(a)/(T) Since the wave velocity (dx)/(dt) is negative, the wave is travelling in the negative x -direction.

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A wave travelling on a string stretched along the X -axis has an...

A wave travelling on a string stretched along the

X

-axis has an equation given by

y = A\, e^{-\left(\dfrac{x}{a} + \dfrac{t}{T}\right)^2}

. Identify the direction in which the wave is travelling.

Options

APositive $y$ -direction
BNegative $x$ -direction
CNegative $y$ -direction
DPositive $x$ -direction

Correct answer

B. Negative $x$ -direction

Step-by-step solution

The given equation of the wave is

y = A e^{-\left(\dfrac{x}{a} + \dfrac{t}{T}\right)^2}

.

For a point on the wave with a constant phase, we have:

\dfrac{x}{a} + \dfrac{t}{T} = \text{constant}

Differentiating with respect to time

t

, we get:

\dfrac{1}{a} \dfrac{dx}{dt} + \dfrac{1}{T} = 0

\dfrac{dx}{dt} = -\dfrac{a}{T}

Since the wave velocity

\dfrac{dx}{dt}

is negative, the wave is travelling in the negative

x

-direction.

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