A particle executes S.H.M with amplitude of 10cm and period of 10s. Find the velocity and acceleration of the particle at a distance 5cm from the equilibrium position

To find the velocity and acceleration of the particle at a distance of 5cm from the equilibrium position, we can use the equations of simple harmonic motion (SHM).

First, let's define some variables:
- Amplitude (A) = 10cm
- Period (T) = 10s
- Equilibrium position (x) = 0cm
- Distance from equilibrium position (d) = 5cm

1. Velocity:
The velocity of the particle in SHM can be determined by the equation: v = ω * A * sin(ωt + φ), where ω is the angular frequency and φ is the phase constant.

The angular frequency (ω) is given by the formula: ω = 2π / T, where T is the period.

Substituting the values, we have:
ω = 2π / 10s
ω = π / 5s

To find the phase constant (φ), we need to determine the phase of the particle at the given distance of 5cm. At the equilibrium position (x = 0cm), the phase is 0. Since the particle is 5cm away from the equilibrium position, the phase would be halfway through its motion.

At the halfway point, the phase is π radians. Therefore, φ = π.

Now we can calculate the velocity:
v = π / 5s * 10cm * sin(π/5 * t + π)

To find the velocity at a distance of 5cm, substitute d = 5cm into the equation:
5cm = 10cm * sin(π/5 * t + π)

Solving for t:
sin(π/5 * t + π) = 0.5
π/5 * t + π = sin^(-1)(0.5)
π/5 * t + π = π/6 or π - π/6
π/5 * t = π/6 - π or π - π/6 - π
π/5 * t = -π/6 or 5π/6

Solving for t:
t = -6/5 or 6

Substituting the value of t into the velocity equation:
v = π / 5s * 10cm * sin(π/5 * t + π)
v = π / 5s * 10cm * sin(π/5 * 6 + π)

Calculate the value using a calculator:
v ≈ 12.57 cm/s

Therefore, the velocity of the particle at a distance of 5cm from the equilibrium position is approximately 12.57 cm/s.

2. Acceleration:
The acceleration of the particle in SHM can be determined by the equation: a = -ω^2 * A * sin(ωt + φ)

Substituting the values, we have:
a = -(π / 5s)^2 * 10cm * sin(π/5 * t + π)
a = -π^2 / 25s^2 * 10cm * sin(π/5 * t + π)

Calculating the value using a calculator:
a ≈ -98.77 cm/s^2

Therefore, the acceleration of the particle at a distance of 5cm from the equilibrium position is approximately -98.77 cm/s^2.

y = 0.10 sin (2 pi * t/10) = 0.10 sin( pi t/5)

v = 0.10 (pi/5) cos (pi t/5)
a = -0.10 (pi/5)^2 sin(pi t/5) = -(pi/5)^2 y
now
find t for y = 0.05
use that t for your v and a