A particle performing SHM has time period 2Π/√3 and path length 4cm. The displacement from mean position at which acceleration is equal to velocity a. Zero b. 0.5cm c. 1cm d. 1.5cm?

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Can anyone have explaination..??

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1cm...

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Hey...what's the answer???mine is coming 2...

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Ans is c.1cm

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Given:
Time period (T) = 2π/√3
Path length (s) = 4 cm

To Find:
The displacement from the mean position at which acceleration is equal to velocity.

Solution:
Step 1: Understanding Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts towards the mean position. It can be represented by an oscillating particle attached to a spring.

Step 2: Relationship between Time Period and Path Length
The time period (T) of an SHM is the time taken for one complete cycle of oscillation. It is related to the path length (s) by the formula:
T = s/v
where v is the velocity of the particle.

Step 3: Relationship between Velocity and Displacement
The velocity (v) of a particle in SHM can be represented as:
v = ω√(A^2 - x^2)
where ω is the angular frequency and A is the amplitude of oscillation. x represents the displacement of the particle from the mean position.

Step 4: Relationship between Acceleration and Displacement
The acceleration (a) of a particle in SHM can be represented as:
a = ω^2x

Step 5: Finding the Displacement
We are given the time period (T) as 2π/√3. From the relationship between time period and path length, we have:
T = s/v
2π/√3 = 4/v

Solving for v, we get:
v = (4√3)/2π
v = √3/π

Now, we know that acceleration (a) is equal to velocity (v) at a certain displacement (x). Therefore, we can equate the two equations:
ω^2x = v
(2π/T)^2x = √3/π

Simplifying the equation, we get:
4π^2x = √3
x = √3/(4π^2)

Calculating the value of x, we find:
x ≈ 0.054 cm

Therefore, the displacement from the mean position at which acceleration is equal to velocity is approximately 0.054 cm, which is closest to option (a) 'Zero'.

Conclusion:
The displacement from the mean position at which acceleration is equal to velocity is approximately 0.054 cm, which is closest to option (a) 'Zero'.

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