Assertion (A): The phase difference between displacement and velocity in SHM is $9 0^{\circ}$
Reason (R): The displacement is represented by y=A sin $ω$ t and Velocity by V=A $ω$ cos $ω$ t.

Question

Verified by Toppr

Answer 1

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero.
Thus displacement can be represented by
$y$ = $A s i n w t$
$V = \frac{d y}{d t}$

$= A w c o s w t = A w s i n (w t + \frac{π}{2})$
Thus phase difference is $\frac{π}{2}$

Answer 2

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero.
Thus displacement can be represented by

y

=

A s i n w t

V = \frac{d y}{d t}

= A w c o s w t = A w s i n (w t + \frac{π}{2})

Thus phase difference is

\frac{π}{2}

Assertion (A): The phase difference between displacement and velocity in SHM is $9 0^{\circ}$
Reason (R): The displacement is represented by y=A sin $ω$ t and Velocity by V=A $ω$ cos $ω$ t.

Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not the correct explanation of A

A is true and R is false

A is false and R is true

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero.
Thus displacement can be represented by
$y$ = $A s i n w t$
$V = \frac{d y}{d t}$

$= A w c o s w t = A w s i n (w t + \frac{π}{2})$
Thus phase difference is $\frac{π}{2}$

A particle is executing SHM $x = 3 cos ω t + 4 sin ω t$ . Find the phase shift and amplitude.

The phase shift of the resultant $y = 3 sin ω t + 4 cos ω t$ is:

What is the phase difference between two simple harmonic motions represented by $x_{1} = A sin (ω t + \frac{π}{6})$ and $x_{2} = A cos (ω t)$ ?

The phase difference between displacement and acceleration of a particle executing simple harmonic motion is:

The phase difference between the velocity and displacement of a particle executing SHM is

Revise with Concepts

Velocity and Acceleration in SHM

Graphical Representation of Displacement, Velocity and Acceleration

Assertion (A): The phase difference between displacement and velocity in SHM is 90∘ Reason (R): The displacement is represented by y=A sinωt and Velocity by V=Aωcosωt.

Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not the correct explanation of A

A is true and R is false

A is false and R is true

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero. Thus displacement can be represented by y= A sin wt V=dtdy​ =Awcoswt=Awsin(wt+2π​) Thus phase difference is 2π​

A particle is executing SHM x=3cosωt+4sinωt. Find the phase shift and amplitude.

The phase shift of the resultant y=3sinωt+4cosωt is:

What is the phase difference between two simple harmonic motions represented by x1​=Asin(ωt+6π​) and x2​=Acos(ωt)?

The phase difference between displacement and acceleration of a particle executing simple harmonic motion is:

The phase difference between the velocity and displacement of a particle executing SHM is

Revise with Concepts

Velocity and Acceleration in SHM

Graphical Representation of Displacement, Velocity and Acceleration

Assertion (A): The phase difference between displacement and velocity in SHM is $9 0^{\circ}$
Reason (R): The displacement is represented by y=A sin $ω$ t and Velocity by V=A $ω$ cos $ω$ t.

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero.
Thus displacement can be represented by
$y$ = $A s i n w t$
$V = \frac{d y}{d t}$

$= A w c o s w t = A w s i n (w t + \frac{π}{2})$
Thus phase difference is $\frac{π}{2}$

A particle is executing SHM $x = 3 cos ω t + 4 sin ω t$ . Find the phase shift and amplitude.

The phase shift of the resultant $y = 3 sin ω t + 4 cos ω t$ is:

What is the phase difference between two simple harmonic motions represented by $x_{1} = A sin (ω t + \frac{π}{6})$ and $x_{2} = A cos (ω t)$ ?