A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y isa)A√3/2b)2A/√3c)A/2d)A/√2Correct answer is option 'A'. Can you explain this answer?

Question

A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y isa)A&#8730;3/2b)2A/&#8730;3c)A/2d)A/&#8730;2Correct answer is option 'A'. Can you explain this answer?

Answer

The relation between angular frequency and displacement is given as
v=ω√A²−x²
Suppose
x=A sinω t
On differentiating the above equation w.r.t. time we get
dx/dt=Aωcosωt
The maximum value of velocity will be [{v{\max }} = A\omega \]
The displacement for the time when speed is half the maximum is given as
v=Aω/2
A²ω²=4ω(A²−x²)
By substituting the value in (1) we get the displacement as
x=A√3/2

Answer

Answer

Explanation:

To solve this problem, we need to use the concepts of Simple Harmonic Motion (SHM) and the relationship between speed and displacement in SHM.

Simple Harmonic Motion (SHM):
Simple Harmonic Motion is a type of periodic motion in which an object oscillates back and forth around an equilibrium position. It can be described by an equation of the form:

y = A * sin(ωt + φ)

where:
- y is the displacement of the particle from its equilibrium position,
- A is the amplitude of the motion,
- ω is the angular frequency (2π divided by the time period T),
- t is the time,
- φ is the phase angle.

Relationship between speed and displacement in SHM:
The speed of a particle in SHM is given by the derivative of the displacement equation with respect to time:

v = dy/dt = A * ω * cos(ωt + φ)

The maximum speed (v_max) occurs when the cosine term is equal to 1, so:

v_max = A * ω

To find the time when the speed is half of the maximum speed, we can set the speed equation equal to half of the maximum speed:

v_half = (1/2) * v_max
v_half = (1/2) * A * ω

Finding the displacement at the time when the speed is half of the maximum speed:
To find the displacement at this time, we need to substitute the value of v_half into the speed equation and solve for y:

(1/2) * A * ω = A * ω * cos(ωt + φ)

Canceling out A and ω:

(1/2) = cos(ωt + φ)

Since the cosine function is equal to 1/2 when the angle is π/3, we can set ωt + φ equal to π/3:

ωt + φ = π/3

Now we need to find the value of y when ωt + φ = π/3. Substituting this value into the displacement equation:

y = A * sin(π/3) = A * (√3/2)

Simplifying:

y = A * (√3/2) = (A√3)/2

Therefore, the correct answer is option 'A': y = (A√3)/2.

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