A vegetable seller in a sabzi mandi hangs a bag of onions on a spring balance. The pointer does not settle at once. It jumps past the reading, swings back above it, and bounces a few times before it settles at \(2\,\text{kg}\). Watch closely: the bounces are centred on \(2\,\text{kg}\), not on zero, where the pointer started.
The model oscillator
A spring-block system is a block fixed to one end of a light spring, whose other end is held still. Lay it on a smooth horizontal floor. Along the floor, the only force on the block is the spring force. Its weight and the normal force from the floor act vertically and cancel. At the natural length of the spring the net force is zero, so that is the mean position.
Apply the two steps. Pull the block a distance \(x\) from the mean position. Step 1: by Hooke's law the spring pulls back with \(F = -kx\), so \(k_{\text{eff}}\) is simply the spring constant \(k\). Step 2: the force moves the block alone, since the spring is light, so \(m_{\text{eff}} = m\). The block therefore moves in SHM with \(\omega = \sqrt{k/m}\).
Watch the point on the graph in Fig 2.1. It never leaves the straight line. When the block is right of the mean, the point is in the lower right quarter: \(x\) is positive and \(F\) negative. When the block is left of the mean, the point is in the upper left quarter. At the ends of the swing the point reaches the ends of the line, \(\pm 4\,\text{N}\). The period is \(2\pi\sqrt{2/20} = 1.99\,\text{s}\). Every oscillator in this chapter will draw a line like this one; only its slope and its body will change.
The line through the origin is the signature of SHM. A restoring force that bends away from the line still gives an oscillation, as for a pendulum at large angles. But then the period starts to depend on the amplitude.
Why the floor must be smooth
Friction changes the story. Kinetic friction has a fixed size and always opposes the velocity, not the displacement. It is not part of \(-kx\). Each swing it turns some of the energy into heat, so the swings shrink and the block stops. The chapter Superposition, Damped and Forced Oscillations studies such damped motion. For SHM that lasts for ever, the floor must be smooth.
A block oscillates on a horizontal spring. For ideal SHM that never fades, what must the coefficient of friction between the block and the floor be?
A steady push moves the centre, not the clock
Now let a steady force \(F_0\) act on the block as well, along the spring. A wind of constant strength, a string over a pulley with a weight, or gravity along a slope would do. The net force becomes \(-kx + F_0\). It is no longer zero at \(x = 0\). It is zero at \(x_0 = F_0/k\), where the spring force balances the steady force. That point is the new equilibrium.
Measure the displacement from the new equilibrium, \(X = x - x_0\). The net force is \(-kx + F_0 = -k(x - F_0/k) = -kX\). The steady force has vanished from the law. About the new mean the block feels the same force constant \(k\) as before, and it has the same mass. So it oscillates about \(x_0\) with the same period.
\[ m\dfrac{d^2x}{dt^2} = -kx + F_0 \]
Why. Newton's second law with the spring force and the steady force together.
\[ x_0 = \dfrac{F_0}{k}, \qquad X = x - x_0 \]
Why. the net force is zero at \(x_0\); measure the displacement \(X\) from there.
\[ m\dfrac{d^2X}{dt^2} = -kX \]
Why. \(x_0\) is a constant, so \(d^2X/dt^2 = d^2x/dt^2\), and \(-kx + F_0 = -k(x - x_0)\).
\[ T = 2\pi\sqrt{\dfrac{m}{k}} \]
Read it. the same law as before, now in \(X\). The motion is SHM about \(x_0\) with the old period.
Let a constant force \(F_0\) act on a body that has a linear restoring force \(-kx\). The body then performs SHM about a new mean position \(x_0 = F_0/k\), with the same period \(2\pi\sqrt{m/k}\).
A \(0.5\,\text{kg}\) block on a smooth floor is fixed to a wall by a spring of constant \(200\,\text{N/m}\). A thread over a pulley applies a steady pull of \(6\,\text{N}\) away from the wall. Find the new mean position and the period.
The new mean is where the spring balances the pull: \(x_0 = F_0/k = 6/200 = 0.03\,\text{m} = 3\,\text{cm}\) beyond the natural length.
The period is \(T = 2\pi\sqrt{0.5/200} = 0.314\,\text{s}\), exactly as without the pull.
If the block is released at the natural length, it starts \(3\,\text{cm}\) from the new mean. It then swings between the natural length and \(6\,\text{cm}\) beyond it.
A particle is in SHM under a restoring force \(-kx\). A constant force is then applied to it as well. What happens?
A \(2\,\text{kg}\) block rests on a smooth incline of \(30°\), held by a spring of constant \(200\,\text{N/m}\) fixed at the top. Take \(g = 10\,\text{m/s}^2\). The block is pulled a little down the slope and released. Which statement is true?
The spring hung vertically
Hang the block from a spring fixed to a ceiling. Now its weight \(mg\) acts along the spring as well, a constant force. By the principle above, the weight moves the mean position and leaves the period alone. The new mean is where the spring balances the weight. There the spring is stretched by \(e\), with \(ke = mg\). We call it the static extension.
Follow the readouts of Fig 2.2. At the lowest point the spring is stretched by \(0.55\,\text{m}\) and pulls up with \(13.75\,\text{N}\). The weight is \(10\,\text{N}\), so the net force is \(3.75\,\text{N}\) upward. At the highest point the spring is stretched by \(0.25\,\text{m}\) and pulls up with only \(6.25\,\text{N}\): the net force is \(3.75\,\text{N}\) downward. At the new mean the two forces balance. The net force is always \(-ky\), with \(y\) measured from the stretched position.
The period is \(2\pi\sqrt{m/k} = 2\pi\sqrt{1/25} = 1.26\,\text{s}\). There is a second way to write it. From \(ke = mg\), the ratio \(m/k\) equals \(e/g\). So the period of a hanging mass can be read from its static extension alone.
The formula \(2\pi\sqrt{e/g}\) does not mean that the period depends on gravity. On the Moon a smaller \(g\) gives a smaller \(e\) as well, and the ratio \(e/g = m/k\) is unchanged. The period depends only on \(m\) and \(k\).
A \(0.5\,\text{kg}\) mass hung on a spring stretches it by \(10\,\text{cm}\). Take \(g = 10\,\text{m/s}^2\). The mass is pulled down a little and released. Find the spring constant and the period.
At rest the spring balances the weight: \(ke = mg\), with \(e = 0.10\,\text{m}\).
\(k = mg/e\), and \(T = 2\pi\sqrt{e/g}\).
\(k = 0.5 \times 10/0.10 = 50\,\text{N/m}\). \(T = 2\pi\sqrt{0.10/10} = 2\pi \times 0.1 = 0.628\,\text{s}\).
The mass bobs about \(1.6\) times a second, about the point \(10\,\text{cm}\) below the natural length.
A load hung on a spring stretches it by \(2.5\,\text{cm}\) at rest. Take \(g = 10\,\text{m/s}^2\). What is the period of its vertical oscillation, in seconds?
A block hanging from a spring oscillates with a period of \(1\,\text{s}\) on the Earth. The same block and spring are taken to the Moon, where \(g\) is about one sixth as large. What is the period there?
A \(0.5\,\text{kg}\) ball hangs from a rubber cord of constant \(50\,\text{N/m}\). A rubber cord pulls only when stretched; it goes slack when compressed. Take \(g = 10\,\text{m/s}^2\). What is the largest amplitude, in centimetres, for which the ball moves in SHM?
The spring balance of the opening is a vertical spring. The weight of the bag shifts the mean position to the \(2\,\text{kg}\) mark. So the pointer swings about that mark, not about zero. The period depends only on the mass of the bag and the stiffness of the spring. It would be the same in any market, on a mountain or on the Moon. On the Moon only the reading itself would be six times smaller.
When six passengers crowd into an auto-rickshaw, the body sinks on its springs. By \(T = 2\pi\sqrt{e/g}\), a larger sag means a slower bounce. A car that sinks \(10\,\text{cm}\) under its load bounces with a period of about \(0.63\,\text{s}\), roughly one and a half bounces a second. Designers of vehicle suspensions choose springs that give about one or two bounces a second. Passengers find this rhythm, close to that of walking, comfortable.
- Taking the natural length as the mean position of a hanging block.The mean is where the spring balances the weight, a static extension \(e = mg/k\) below the natural length.
- Thinking that the weight changes the period of a vertical spring.The weight is a constant force. It moves the mean position and leaves \(T = 2\pi\sqrt{m/k}\) unchanged.
- Writing the spring force of a hanging block as \(ky\), with \(y\) from the mean.The spring force is \(k(e - y)\) with the full stretch. It is the net force, spring and weight together, that equals \(-ky\).