A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t=0 it is at position x=5 cm going towards positive x direction . Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4s.

A=10cm

w=2PIf=2PI/T= PI/3

f(t)=Acos(wt+Phi)

f(0)=10cm*cos(0+phi)=5 >>>phi=PI/3 radians

f(t)=10cm*cos(wt+PI/3)
f'=-10cm*w*sin(wt+PI/3)=velocity
f"=-10cm*w^2*cos(PI*t/3+PI/3)=acceleration

Q 2) A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

X=(10cm)sin(2¦Ð/6st+¦Ð/6),~11cm s

To write the equation for the displacement x at time t, we can use the general equation for simple harmonic motion:

x = A * cos(ωt + ϕ)

Where:
- x is the displacement of the particle from its equilibrium position
- A is the amplitude of the motion (given as 10 cm)
- ω is the angular frequency of the motion, which can be calculated using the formula ω = 2π / T, where T is the time period (given as 6 s)
- t is the time at which we want to find the displacement
- ϕ is the phase constant, representing the initial phase of the motion

Now, let's calculate ω:

ω = 2π / T
= 2π / 6
= π / 3 rad/s

Substituting the given values, we have:

x = 10 * cos((π / 3)t + ϕ)

To find the value of ϕ, we need additional information. The problem states that at t=0, the particle is at position x=5 cm, moving towards the positive x direction. Since the cos function reaches its maximum value of 1 when the argument is 0, we can infer that at t=0, ϕ is 0.

Therefore, the equation for the displacement x at time t is:

x = 10 * cos((π / 3)t)

To find the magnitude of the acceleration at t=4s, we need to differentiate the displacement equation with respect to time twice. The second derivative of x with respect to time will give us the acceleration:

a = -A * ω^2 * sin(ωt + ϕ)

Let's calculate the acceleration at t=4s:

t = 4s
ω = π / 3 rad/s

a = -10 * (π / 3)^2 * sin((π / 3) * 4 + 0)

The magnitude of the acceleration can be found by taking the absolute value of the result:

|a| = |-10 * (π / 3)^2 * sin((π / 3) * 4)|
≈ |-10 * (3.2899)|
≈ 32.899 cm/s^2

Therefore, the magnitude of the acceleration of the particle at t=4s is approximately 32.899 cm/s^2.