
Find the maximum transverse velocity and acceleration of particles in the wire.
Answer
518.1k+ views
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.
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.
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