Answer:
Therefore, the velocity of the particle when it is at a distance of 2cm from its mean position is -4√3 * sin(√3t) cm/s.
Step-by-step explanation:
To determine the velocity of a particle executing Simple Harmonic Motion (SHM) at a distance of 2cm from its mean position, we can use the equations of motion for SHM.
In SHM, the displacement of the particle from its mean position can be represented by the equation:
x = A * cos(ωt + φ)
Where:
- x is the displacement of the particle
- A is the amplitude of the motion (maximum displacement)
- ω is the angular frequency of the motion
- t is the time
- φ is the phase constant
The acceleration of the particle at any point in SHM can be given by:
a = -A * ω^2 * sin(ωt + φ)
Given that the maximum displacement (amplitude) is 4cm, we have A = 4cm. We are also given that the acceleration at a distance of 1cm from the mean position is 3cm/s^2. Therefore, when x = 1cm, a = 3cm/s^2.
Substituting these values into the equation for acceleration, we get:
3 = -4 * ω^2 * sin(ωt + φ)
Now, let's find ω using the given information. At x = 1cm, we know that a = -ω^2 * x. Substituting x = 1cm and a = 3cm/s^2, we have:
3 = -ω^2 * 1
ω^2 = -3
Since ω represents angular frequency, which must be positive, we take the positive square root:
ω = √3 rad/s
Now, let's find φ. At t = 0s, x = A * cos(φ). Substituting x = 4cm and A = 4cm, we have:
4 = 4 * cos(φ)
cos(φ) = 1
φ = 0
Therefore, the equation for displacement becomes:
x = 4 * cos(√3t)
To find the velocity at x = 2cm, we need to differentiate the displacement equation with respect to time:
v = dx/dt = -4√3 * sin(√3t)
Now, let's find the velocity when x = 2cm. Substituting x = 2cm into the velocity equation, we have:
v = -4√3 * sin(√3t) = -4√3 * sin(√3t) cm/s