AP Physics 1 · Oscillations · free lesson

Defining Simple Harmonic Motion (SHM)

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Periodic motion and equilibrium

A swing, a guitar string, a mass bouncing on a spring and the pendulum of a clock all move back and forth, over and over, along the same path. Motion that repeats itself at regular intervals of time is called periodic motion. The Moon going around Earth is periodic too, and so is a ball bouncing on a floor, but they are not the kind of periodic motion that this topic is about.

Definition.

Simple harmonic motion (SHM) is a special case of periodic motion. An object moves in simple harmonic motion when the restoring force on it has a magnitude proportional to its displacement from its equilibrium position.

Two words in this definition need to be clear before the rule can be used.

Definition.

An equilibrium position is a location at which the net force exerted on an object or system is zero. An object placed there at rest stays there.

Definition.

A restoring force is a force exerted in a direction opposite to the object's displacement from an equilibrium position. It always pushes or pulls the object back toward equilibrium.

The displacement here is measured from the equilibrium position, not from some other origin: it is zero at equilibrium, positive on one side and negative on the other. A ball on the floor of a bowl shows the idea. At the lowest point the net force is zero. If the ball is pushed up one side, the net force on it points back down the slope, and the farther up it is, the steeper the slope and the bigger the force.

The condition for simple harmonic motion

Rule.

SHM results when the magnitude of the restoring force is proportional to the displacement from equilibrium. For a displacement $x$ from the equilibrium position, the force component is $F_x = -kx$ , where $k$ is a positive constant. The minus sign says that the force is opposite to the displacement. Newton's second law then gives the acceleration: $a_x = \dfrac{F_x}{m} = -\dfrac{k}{m}\,x$ .

The ideal spring of topic 2.8 is the standard example: its force is $F_s = k\lvert\Delta x\rvert$ and points toward the relaxed position, so a block on a smooth table attached to a spring has exactly this restoring force, with $k$ the spring constant and the displacement measured from the relaxed position. Both the force and the acceleration are largest at the ends of the motion, where the displacement is greatest, they are zero at the equilibrium position, and they always point toward the equilibrium position.

The animation shows a $0.5\,\mathrm{kg}$ block on a smooth table, joined to a wall by a spring of $40\,\mathrm{N/m}$ . It is released from rest $0.15\,\mathrm{m}$ to the right of the dashed line, which marks the equilibrium position, and goes back and forth. The red arrow is the force of the spring on the block. Watch its length and its direction: it points toward the dashed line on both sides, it is longest at the ends and it shrinks to nothing as the block passes the dashed line.

In the picture the positive direction is to the right, so at the right end the force component is negative, and at the left end it is positive: the sign of the force is always the opposite of the sign of the displacement.

\begin{aligned} F_{max} &= kA \\ &= 40\,\mathrm{N/m} \times 0.15\,\mathrm{m} \\ &= 6.00\,\mathrm{N} \\ a_{max} &= \frac{F_{max}}{m} \\ &= \frac{6\,\mathrm{N}}{0.5\,\mathrm{kg}} \\ &= 12.0\,\mathrm{m/s^{2}} \end{aligned}

At the ends, where the displacement is $0.15\,\mathrm{m}$ , the force is $6.00\,\mathrm{N}$ and the acceleration is $12.0\,\mathrm{m/s^{2}}$ , both toward the equilibrium position. At the dashed line both are zero. Doubling the displacement would double each of them.

The graph of the restoring force

A graph of the force component against the displacement shows at once whether the force is the kind that gives SHM. The graph must be a straight line through the origin with a negative slope: the force is zero at zero displacement, and it has the opposite sign to the displacement. The slope of the line is $-k$ .

\begin{aligned} F_x &= kx \\ &= 40\,\mathrm{N/m} \times 0.1\,\mathrm{m} \\ &= 4.00\,\mathrm{N} \end{aligned}

The graph is for a spring of $40\,\mathrm{N/m}$ . At a displacement of $0.1\,\mathrm{m}$ to the right the force is $4.00\,\mathrm{N}$ to the left, so the point is at $(0.10, -4.0)$ , and at a displacement of $0.1\,\mathrm{m}$ to the left the force is the same size to the right. A restoring force that was the same size at every displacement, or that grew with the square of the displacement, would give a line that is not straight, and the motion would be periodic but not simple harmonic.

Both curves above are restoring forces, because each one has the opposite sign to the displacement everywhere. Only the blue straight line is a force for SHM. The red curve gives a force proportional to the square of the displacement, so doubling the displacement makes the force four times as large. That motion repeats, but it is not simple harmonic.

A mass hanging from a spring

The equilibrium position is where the net force is zero, and that is not always where the spring has its relaxed length. A mass hanging at rest from a vertical spring is in equilibrium when the spring is stretched so that its force balances the weight. Push the mass down by a distance $y$ from there, or lift it by $y$ , and the net force is no longer zero.

\begin{aligned} x_e &= \frac{mg}{k} \\ &= \frac{0.4\,\mathrm{kg} \times 9.8\,\mathrm{m/s^{2}}}{50\,\mathrm{N/m}} \\ &= 0.0784\,\mathrm{m} \\ F_g &= mg \\ &= 0.4\,\mathrm{kg} \times 9.8\,\mathrm{m/s^{2}} \\ &= 3.92\,\mathrm{N} \\ F_s &= k(x_e + y) \\ &= 50\,\mathrm{N/m} \times (0.0784\,\mathrm{m} + 0.06\,\mathrm{m}) \\ &= 6.92\,\mathrm{N} \\ F_{net} &= F_s - F_g \\ &= 6.92\,\mathrm{N} - 3.92\,\mathrm{N} \\ &= 3.00\,\mathrm{N} \end{aligned}

For a $0.4\,\mathrm{kg}$ mass on a spring of $50\,\mathrm{N/m}$ , the equilibrium stretch is $0.0784\,\mathrm{m}$ . Pulled a further $0.06\,\mathrm{m}$ down, the spring force is $6.92\,\mathrm{N}$ up and the weight is $3.92\,\mathrm{N}$ down, so the net force is $3.00\,\mathrm{N}$ upward. It is the same as $ky$ = $3.00\,\mathrm{N}$ : the weight cancels against the part of the spring force that is there at equilibrium, and what is left is $-ky$ , measured from the equilibrium position. A mass hanging on a spring is therefore in SHM about its hanging position, in just the way a block on a table is in SHM about the relaxed position.

The picture shows the mass at three positions with the net force as the red arrow. The block is lower than the dashed line on the left, and the net force is up. On the right it is higher, and the net force is down. In the middle the net force is zero. The drawing is not to scale.

The pendulum

A simple pendulum is a small heavy bob on a light string, swinging about a fixed point. Its equilibrium position is hanging straight down. Two forces act on the bob: its weight $mg$ and the tension $F_T$ of the string. The picture shows the bob displaced by an angle $\theta$ from the vertical. Drag the slider to change the angle.

The blue arrow is the weight and the red arrow is the tension. The tension is drawn shorter than the weight because it is $mg\cos\theta$ when the bob is momentarily at rest at the end of its swing. These two forces do not add up to zero when $\theta$ is not zero: the net force on the bob at that moment is along the path and points back toward the vertical.

To see how big that restoring effect is, split the weight into a part along the string and a part along the path of the bob. This is a separate picture, since a free-body diagram shows whole forces only. The part along the string, $mg\cos\theta$ , is balanced by the tension. The part along the path, $mg\sin\theta$ , is the restoring force.

The green dashed arrow is the part of the weight along the path, the red dashed arrow the part along the string. Torque is the quantity that turns something about a fixed point. The tension acts along the string, which passes through the pivot, so it has no torque about the pivot. The weight's torque about the pivot is the weight's part along the path times the length $L$ of the string.

Rule.

The pendulum. When the bob is displaced to one side, the torque of its weight about the pivot is a restoring torque, opposite to the angular displacement: $\tau = -mgL\sin\theta$ . For a small angular displacement (in radians) $\sin\theta \approx \theta$ , so $\tau \approx -(mgL)\,\theta$ . The restoring torque is then proportional to the angular displacement, and the pendulum can be modeled as simple harmonic motion.

This is the same pattern as the spring: a restoring effect proportional to the displacement, with the angular displacement taking the place of $x$ . For a larger swing the approximation fails, and the motion is no longer simple harmonic, only periodic. The graph compares $\sin\theta$ with $\theta$ in radians.

The blue line is $\theta$ and the red curve is $\sin\theta$ . They are almost the same for small angles and move apart as the angle grows. At 5 degrees ( $0.0873$ rad) the sine is $0.0872$ , which differs from the angle by $0.13$ percent. At 20 degrees ( $0.349$ rad) the sine is $0.342$ , and the difference is $2.1$ percent. At 45 degrees ( $0.785$ rad) it is $0.707$ and the difference is $11$ percent. This is why the pendulum is treated as SHM for small swings only, of the order of 10 degrees or less.

Warning.

Three traps. The displacement is measured from the equilibrium position, so a hanging mass has its equilibrium where the spring is already stretched. A motion that merely repeats, such as a ball bouncing on the floor, is periodic but is not SHM, because its force is not proportional to a displacement from equilibrium. And a pendulum is SHM only for small angles, because the restoring torque is proportional to $\sin\theta$ and not to $\theta$ .

Worked examples

Five examples of the kinds the AP exam asks on this topic: the force and acceleration of an oscillating block, a force graph read, a hanging mass, a pendulum at a small and a large angle, and a "what changes" question. Use g = 9.8 m/s². Treat every spring as ideal and every surface as smooth.

Example.

Example 1. The force and acceleration of an oscillating block

A $0.6\,\mathrm{kg}$ block on a smooth table is fastened to a spring of $30\,\mathrm{N/m}$ . The block is pulled $0.2\,\mathrm{m}$ to the right of its equilibrium position and released, and it moves back and forth. Find the force on the block and its acceleration at the instant of release, and half-way back to equilibrium. Give the directions.

Solution

The table is smooth, so the spring force is the only horizontal force, and it is the restoring force: its magnitude is $k$ times the displacement from equilibrium, and it points toward equilibrium.

\begin{aligned} F_1 &= k_1A_1 \\ &= 30\,\mathrm{N/m} \times 0.2\,\mathrm{m} \\ &= 6.00\,\mathrm{N} \\ a_1 &= \frac{F_1}{m_1} \\ &= \frac{6\,\mathrm{N}}{0.6\,\mathrm{kg}} \\ &= 10.0\,\mathrm{m/s^{2}} \\ F_h &= \frac{k_1A_1}{2} \\ &= \frac{30\,\mathrm{N/m} \times 0.2\,\mathrm{m}}{2} \\ &= 3.00\,\mathrm{N} \\ a_h &= \frac{F_h}{m_1} \\ &= \frac{3\,\mathrm{N}}{0.6\,\mathrm{kg}} \\ &= 5.00\,\mathrm{m/s^{2}} \end{aligned}

Result.

At release the force is $6.00\,\mathrm{N}$ and the acceleration is $10.0\,\mathrm{m/s^{2}}$ , both to the left, toward equilibrium. Half-way back, at a displacement of $0.1\,\mathrm{m}$ , both are half as large: $3.00\,\mathrm{N}$ and $5.00\,\mathrm{m/s^{2}}$ , still to the left. At the equilibrium position both are zero. The acceleration is not constant, so the equations for constant acceleration cannot be used for this motion.

Example.

Example 2. Reading a graph of force against displacement

The graph shows the force component on a block against its displacement from equilibrium. Show that the block can be in simple harmonic motion, find the constant $k$ , and find the magnitude of the acceleration when the block's displacement is $0.08\,\mathrm{m}$ and its mass is $0.5\,\mathrm{kg}$ .

Solution

The graph is a straight line through the origin, and the force has the opposite sign to the displacement: the force is a restoring force proportional to the displacement, which is the condition for SHM. The red point is at x = $0.08\,\mathrm{m}$ , F = $-$ $2.00\,\mathrm{N}$ . The slope of the line is $-k$ .

\begin{aligned} k_g &= \frac{F_2}{x_2} \\ &= \frac{2\,\mathrm{N}}{0.08\,\mathrm{m}} \\ &= 25.0\,\mathrm{N/m} \\ a_2 &= \frac{k_gx_2}{m_2} \\ &= \frac{25\,\mathrm{N/m} \times 0.08\,\mathrm{m}}{0.5\,\mathrm{kg}} \\ &= 4.00\,\mathrm{m/s^{2}} \end{aligned}

Result.

The constant is $25.0\,\mathrm{N/m}$ , the size of the slope. The acceleration has magnitude $4.00\,\mathrm{m/s^{2}}$ , directed opposite to the displacement. A line that did not pass through the origin, or that had a curve in it, would not describe SHM.

Example.

Example 3. A hanging mass

A $0.3\,\mathrm{kg}$ mass hangs at rest from a vertical spring of $60\,\mathrm{N/m}$ . It is pulled down a further $0.04\,\mathrm{m}$ and released. Find the net force and the acceleration on the mass at the instant of release, and compare them with $ky$ .

Solution

First find where the equilibrium position is: the spring is stretched until its force equals the weight. Then add the extra stretch.

\begin{aligned} F_g &= m_3g \\ &= 0.3\,\mathrm{kg} \times 9.8\,\mathrm{m/s^{2}} \\ &= 2.94\,\mathrm{N} \\ x_e &= \frac{F_g}{k_3} \\ &= \frac{2.94\,\mathrm{N}}{60\,\mathrm{N/m}} \\ &= 0.0490\,\mathrm{m} \\ F_s &= k_3(x_e + y_3) \\ &= 60\,\mathrm{N/m} \times (0.049\,\mathrm{m} + 0.04\,\mathrm{m}) \\ &= 5.34\,\mathrm{N} \\ F_n &= F_s - F_g \\ &= 5.34\,\mathrm{N} - 2.94\,\mathrm{N} \\ &= 2.40\,\mathrm{N} \\ a_3 &= \frac{F_n}{m_3} \\ &= \frac{2.4\,\mathrm{N}}{0.3\,\mathrm{kg}} \\ &= 8.00\,\mathrm{m/s^{2}} \end{aligned}

Result.

The equilibrium stretch is $0.0490\,\mathrm{m}$ . At release the spring force is $5.34\,\mathrm{N}$ up and the weight is $2.94\,\mathrm{N}$ down, so the net force is $2.40\,\mathrm{N}$ up and the acceleration is $8.00\,\mathrm{m/s^{2}}$ up. This equals $ky$ = $2.40\,\mathrm{N}$ : the net force is proportional to the displacement from the equilibrium position, not from the relaxed length. Forgetting the weight would give a net force of $5.34\,\mathrm{N}$ , which is the spring force alone. That is wrong: the net force is the spring force minus the weight.

Example.

Example 4. Is a pendulum in simple harmonic motion?

A simple pendulum is pulled aside by $6$ degrees and released, and a second, identical one by $50$ degrees. For which can the motion be modeled as SHM? Justify using the restoring torque.

Solution

The restoring torque of the weight about the pivot is proportional to $\sin\theta$ . SHM needs a torque proportional to the angular displacement $\theta$ in radians, so compare the two.

Result.

For $6$ degrees ( $0.105$ rad) the sine is $0.105$ , and the angle in radians is larger by $0.18$ percent. The torque is proportional to $\theta$ to within that small amount, so the swing can be modeled as SHM. For $50$ degrees ( $0.873$ rad) the sine is $0.766$ , and the angle is larger by $14$ percent. The torque is clearly not proportional to the angular displacement, and the large swing is periodic but not well modeled as SHM.

Example.

Example 5. What changes?

A $0.4\,\mathrm{kg}$ block on a spring of $50\,\mathrm{N/m}$ is at a displacement of $0.06\,\mathrm{m}$ from equilibrium. How does the magnitude of its acceleration change if (a) the displacement is doubled, (b) the mass is doubled, (c) the spring is replaced by one of twice the spring constant?

Solution

Since $a = \frac{kx}{m}$ , the acceleration is proportional to the displacement and to the spring constant and inversely proportional to the mass.

\begin{aligned} a_0 &= \frac{k_5x_5}{m_5} \\ &= \frac{50\,\mathrm{N/m} \times 0.06\,\mathrm{m}}{0.4\,\mathrm{kg}} \\ &= 7.50\,\mathrm{m/s^{2}} \\ a_d &= \frac{k_5 \times 2x_5}{m_5} \\ &= \frac{50\,\mathrm{N/m} \times 2 \times 0.06\,\mathrm{m}}{0.4\,\mathrm{kg}} \\ &= 15.0\,\mathrm{m/s^{2}} \\ a_m &= \frac{k_5x_5}{2m_5} \\ &= \frac{50\,\mathrm{N/m} \times 0.06\,\mathrm{m}}{2 \times 0.4\,\mathrm{kg}} \\ &= 3.75\,\mathrm{m/s^{2}} \\ a_k &= \frac{2k_5x_5}{m_5} \\ &= \frac{2 \times 50\,\mathrm{N/m} \times 0.06\,\mathrm{m}}{0.4\,\mathrm{kg}} \\ &= 15.0\,\mathrm{m/s^{2}} \end{aligned}

Result.

The acceleration starts at $7.50\,\mathrm{m/s^{2}}$ . (a) Doubling the displacement doubles it, to $15.0\,\mathrm{m/s^{2}}$ . (b) Doubling the mass halves it, to $3.75\,\mathrm{m/s^{2}}$ . (c) Doubling the spring constant doubles it, to $15.0\,\mathrm{m/s^{2}}$ . The direction in each case is toward equilibrium.

Practice

Answer AP-style questions on defining simple harmonic motion (shm). Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise defining simple harmonic motion (shm) with new numbers