# Simple Harmonic Motion of a Particle

• HZXAHNLfzjSr
In summary, a particle is oscillating in simple harmonic motion with a period of 4.5 ms and an amplitude of 3.0 cm. At time t = 0, the particle is at the equilibrium position. We can calculate the frequency to be 220 Hz and the angular frequency to be 1.4 x 10^3 rad/s. The maximum speed is 42 m/s and the maximum acceleration is 5.9 x 10^4 m/s^2. To find the speed at t = 1.0 ms, we can use the equation x = x0 sin (omega t) to get the position x(t) as a function of time. Taking the derivative of this function, we
HZXAHNLfzjSr

## Homework Statement

A particle is oscillating in simple harmonic motion with period 4.5 ms and amplitude 3.0 cm. At time t = 0, the particle is at the equilibrium position. Calculate, for this particle:

a) frequency
b) the angular frequency
c) the maximum speed
d) the maximum acceleration
e) the speed at time t = 1.0 ms

None

## The Attempt at a Solution

a) 220 Hz
c) 42 m/s
d) 5.9 x 10^4 m/s^2I am stuck at (e) and just don't know how to do this.
I was thinking to use the Instantaneous Velocity formula (Vins = omega {square root[(amplitude)^2 - (displacement)^2]} but that isn't giving me the correct answer.

You need to write down an expression that gives you the position x(t) as a function of time. Take the derivative to get the velocity v(t) as a function of time. From that you can easily find the speed at t = 1.0 ms.

Thanks, but can you show me the steps?

x = x0 sin (omega t)

If you differentiate this function with respect to time, you'll get the function of Velocity.

Both equations (Instantaneous and function of time) should give you the same answer, so check your workings again.

Last edited:

For part (e), you can use the formula for velocity in simple harmonic motion: v = ω√(A^2 - x^2), where ω is the angular frequency, A is the amplitude, and x is the displacement from equilibrium position.

At t = 1.0 ms, the displacement from equilibrium position is 3.0 cm (since the particle starts at equilibrium position and oscillates back and forth). So, plugging in the values given in the problem, we get:

v = (1.4 x 10^3 rad/s)√[(3.0 cm)^2 - (3.0 cm)^2]

= 0 cm/s

The speed at t = 1.0 ms is 0 cm/s. This makes sense because at the equilibrium position, the particle momentarily stops before changing direction and moving back towards the opposite direction.

I hope this helps!

## 1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) of a particle is a type of periodic motion where the particle moves back and forth along a straight line with a constant amplitude and a specific frequency.

## 2. What is the equation for Simple Harmonic Motion?

The equation for SHM is x(t) = A sin (ωt + φ), where x(t) is the displacement of the particle from its equilibrium position, A is the amplitude, ω is the angular frequency, and φ is the phase constant.

## 3. What is the difference between Simple Harmonic Motion and Uniform Circular Motion?

While both SHM and Uniform Circular Motion are types of periodic motion, the main difference is that SHM occurs along a straight line, while Uniform Circular Motion occurs along a circular path.

## 4. What factors affect the period of Simple Harmonic Motion?

The period of SHM is affected by the mass of the particle, the force acting on the particle, and the stiffness of the spring (if present). These factors determine the frequency of the motion, and therefore, the period.

## 5. How is Simple Harmonic Motion related to the concept of energy?

In SHM, the total mechanical energy (kinetic + potential) of the particle remains constant, as there is no external force acting on it. This is known as the principle of conservation of energy and is applicable to all types of simple harmonic motion.

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