Velocity and acceleration in SHM

How fast, and how fast it changes

A body in SHM keeps changing its speed. It stops at each end and moves fastest at the centre. How fast is it at each moment? We can find out from the reference circle.

The velocity

P moves round the circle of radius $A$ at the angular speed $\omega$. So its speed is $\omega A$, along the tangent. At the phase $\omega t + \phi$, the part of this velocity along the diameter is

This is the velocity of Q, the body in SHM. The minus sign tells the direction. When the phase is between $0$ and $\pi$, P is above the diameter and moving to the left, so $v$ is negative.

The speed is largest when $\sin(\omega t + \phi) = \pm1$. That happens at $x = 0$, the mean position:

At the ends, $x = \pm A$, the sine is zero, and the body stops for an instant.

There is a useful way to find the speed at any position $x$. Since $\cos^2 + \sin^2 = 1$,

The sign depends on whether the body is moving out or coming back.

The acceleration

P has the centripetal acceleration $\omega^2 A$, pointing to the centre of the circle. Its part along the diameter is

This is the key fact about SHM. The acceleration is proportional to the displacement, and always points towards the mean position. The minus sign says that it is opposite to $x$.

The acceleration is largest at the ends, $a_{max} = \omega^2 A$, where the body turns back. It is zero at the mean position, where the speed is largest. In course 132 we find the force that gives this acceleration.

The phases of x, v and a

Look at the three graphs above. The velocity reaches its peak a quarter period before the position does. We say that $v$ leads $x$ by $\pi/2$. The acceleration is always opposite to $x$, so it leads $x$ by $\pi$. When $x$ is largest, $a$ is most negative and $v$ is zero.

Worked example. A body moves in SHM with $A = 0.05\ \text{m}$ and $T = \pi/20\ \text{s}$. Find its largest speed and largest acceleration.

We want $\omega$ first, so $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{\pi/20\ \text{s}} = 40\ \text{rad/s}$. Then $v_{max} = \omega A = 40\ \text{rad/s} \times 0.05\ \text{m} = 2.0\ \text{m/s}$. And $a_{max} = \omega^2 A = (40\ \text{rad/s})^2 \times 0.05\ \text{m} = 80\ \text{m/s}^2$.

The curve of $v$ against $x$ is an oval, an ellipse. The body goes round it once in each oscillation. The upper half is for the body moving towards $+x$, and the lower half for the way back.

Practise with new numbers

Try it in Python

The program prints the position, velocity and acceleration of a body in SHM every quarter period. The last line checks that $a = -\omega^2 x$.

Common mistakes

• Thinking the acceleration is largest where the speed is largest. It is the other way round. At the centre the speed is largest and the acceleration is zero.
• Thinking the body has no acceleration at an end because it stops there. It stops for an instant, but its acceleration is then the largest.
• Using $v = \omega x$. The speed at $x$ is $\omega\sqrt{A^2 - x^2}$, which is smallest where $x$ is largest.
• Losing the minus sign in $a = -\omega^2 x$. The acceleration always points back to the mean position.

Remember these

• $v = -\omega A\sin(\omega t + \phi)$, and $|v| = \omega\sqrt{A^2 - x^2}$.
• $v_{max} = \omega A$, at the mean position. $v = 0$ at the ends.
• $a = -\omega^2 x$. $a_{max} = \omega^2 A$, at the ends. $a = 0$ at the mean position.
• $v$ leads $x$ by $\pi/2$, and $a$ leads $x$ by $\pi$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A body in SHM has an amplitude of $4\ \text{cm}$ and an angular frequency of $5\ \text{rad/s}$. What is its largest speed?

2. Where in its path is a body in SHM when its acceleration is zero? What is its speed then?

3. A body in SHM has $A = 10\ \text{cm}$ and $\omega = 3\ \text{rad/s}$. Find its speed at $x = 6\ \text{cm}$.

4. At one moment a body in SHM is at $x = +A$. Where is it a quarter period later, and what are its velocity and acceleration then?