Question

Find the maximum transverse velocity and acceleration of particles in the wire.

Answer 1

Hint: Assume an equation of wave i.e., y. Velocity is defined as the differentiate of y with respect to t and Acceleration is defined as the differentiate of v with respect to t. For finding the transverse value of velocity and acceleration, we need to take the maximum value of v and a.

Complete answer:
 Let's take the equation of a wave: -
$y = Asin (kx-wt)$
For finding the maximum transverse velocity, we need to differentiate y with respect to time.
$v = \dfrac{dy}{dt}$
$ \implies v = \dfrac{d( Asin (kx-wt))}{dt}$
$\implies v = -Aw cos (kx-wt)$
The maximum amplitude of v is the transverse velocity.
Since cosine function varies from $ 1 to -1$, as v is negative so we take the value of cosine as $-1$. So, Aw is the transverse velocity.
For finding the maximum transverse acceleration, we need to differentiate v with respect to time.
$a= \dfrac{dv}{dt}$
$ \implies a = \dfrac{d(-Aw cos (kx-wt))}{dt}$
$\implies a = -Aw^{2}sin (kx-wt)$
The maximum amplitude of a is the transverse acceleration.
Since sine function varies from $ 1 to -1$,, as a is negative so we take the value of sine as $-1$. So, $ Aw^{2}$ is the transverse acceleration.
Hence, Aw and $ Aw^{2}$ is the transverse velocity and transverse acceleration respectively.

Note: Acceleration is defined as the double differentiation of y with respect to time.
$a = \dfrac{d^{2}y}{dt^{2}}$
$ \implies a = \dfrac{d^{2}(Asin (kx-wt))}{dt^{2}}$
$\implies a = -Aw^{2}sin (kx-wt)$
The sine function varies from $-1 to 1$. Therefore, its greatest value is $1$. Since the angular frequency​ and amplitude do not change, it is this sine that makes velocity vary. If we maximize the sine, we maximize the acceleration. Since the sine's maximum value is $1$, but acceleration is already negative so to make it positive, we have to take the negative value.