Springs, pulleys and moving supports

A spring scale in a lift

A fish hangs on a spring scale. The pointer rests at 2.0 kg. Now the lift you are standing in starts to rise. The pointer swings down past a new reading and comes back, and then it swings again. Why does it swing at all, and about which reading does it swing?

The free courses used a fixed support. A moving support needs new ideas. This lesson has three: a pulley that changes the stiffness, a support that accelerates, and a mass that changes in the middle of a swing.

One method for any spring system

Do not memorise a table of formulas. Use one method.

1. Move the body a small distance $x$ from its balance point.
2. Find the restoring force $F$ on it, or the stored energy $U$.
3. Write $F = -k_{eff}\,x$ or $U = \tfrac12 k_{eff}x^2$. This gives the effective spring constant $k_{eff}$.
4. Then $\omega = \sqrt{\dfrac{k_{eff}}{m}}$ and $T = 2\pi\sqrt{\dfrac{m}{k_{eff}}}$.

You know that springs side by side add: $k_{eff} = k_1 + k_2$. Springs end to end give $\dfrac{1}{k_{eff}} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$. In a network, reduce the series pairs first, then add what is left in parallel. Where a pulley or a lever is present, the method still works. You only have to ask how far the spring stretches for each metre that the body moves.

A worked example. A block of $1.2$ kg hangs from a light movable pulley. A string runs from the ceiling, round the pulley, and up to a spring of $100\ \text{N/m}$. The upper end of the spring is fixed to the ceiling.

1. The constraint. Let the block go down by $y$. The string on each side of the pulley gets longer by $y$. The spring must give $2y$ of string, so it stretches by $2y$.
2. The energy. $U = \tfrac12 k(2y)^2 = \tfrac12(4k)\,y^2$. So $k_{eff} = 4k = 400\ \text{N/m}$.
3. The period. $T = 2\pi\sqrt{\dfrac{1.2\ \text{kg}}{400\ \text{N/m}}} = 0.344$ s. The same block on the spring alone would take $0.688$ s.

A pulley that makes the spring stretch twice as much makes the whole system four times as stiff. Two things add up. The spring moves twice as far for each metre of the block's move, and two strands of string pull on the block, so the force on it is doubled too.

A support that accelerates

Back to the lift. Let the block of mass $m$ rest on a spring of constant $k$ fixed to a trolley. The trolley starts to speed up with an acceleration $a$.

In the frame of the trolley, the block feels an extra force $ma$ in the direction opposite to $a$. The spring force must now balance $ma$, so the middle of the swing moves back by

Nothing else changes. The spring constant and the mass are the same, so the period is unchanged. A vertical spring in a lift works the same way, with $g$ replaced by $g + a$ for the weight. The new middle is $m(g + a)/k$ below the natural length. The shift from the old middle is again $ma/k$.

The block was at rest at the old middle when the acceleration began. It is now a distance $s$ from the new middle, and it is not moving along the trolley. So it swings with amplitude $s$. Its position along the trolley, from the old middle, is

The block goes back as far as $2s$ from the old middle, and its greatest speed along the trolley is $\omega s = a\sqrt{m/k}$.

A worked example. A block of $0.50$ kg is on a spring of $200\ \text{N/m}$ on a trolley. The trolley speeds up at $2.0\ \text{m/s}^2$.

1. The shift. $s = \dfrac{0.50\ \text{kg} \times 2.0\ \text{m/s}^2}{200\ \text{N/m}} = 5.0$ mm.
2. The period. $T = 2\pi\sqrt{\dfrac{0.50\ \text{kg}}{200\ \text{N/m}}} = 0.314$ s.
3. The greatest speed. $\omega s = 20\ \text{rad/s} \times 5.0\ \text{mm} = 0.100\ \text{m/s}$, along the trolley.
4. The farthest point. $2s = 10$ mm behind the old middle.

Make the acceleration $a$ larger. The dashed line moves, but the time for one swing does not. Now make the spring stiffer. What happens to the shift and to the period?

A lump that sticks

A block of mass $M$ on a spring swings on a smooth table with amplitude $A$. A lump of putty of mass $m$ falls straight down onto the block and sticks to it. The putty has no sideways speed. Two things can change: the mass, so the period, and the energy, so the amplitude. It depends on when the putty lands.

At the middle. The block moves at its greatest speed $v = \omega A$. The sideways momentum is conserved, so the new speed is $v' = \dfrac{Mv}{M + m}$. The new angular frequency is $\omega' = \sqrt{\dfrac{k}{M + m}}$. The new amplitude is $A' = \dfrac{v'}{\omega'}$, which gives

Some of the energy turns into heat in the collision.

At the end of the swing. The block is at rest. The putty lands on a block that is not moving, so no energy is lost. The amplitude stays $A$. Only the period changes, to $2\pi\sqrt{(M + m)/k}$.

A worked example. $M = 0.60$ kg, $k = 240\ \text{N/m}$, $A = 0.10$ m and $m = 0.20$ kg. The block's $\omega$ is $20$ rad/s.

1. At the middle. $v = 20\ \text{rad/s} \times 0.10\ \text{m} = 2.0\ \text{m/s}$. Then $v' = \dfrac{0.60 \times 2.0}{0.80}\ \text{m/s} = 1.5\ \text{m/s}$ and $\omega' = \sqrt{\dfrac{240\ \text{N/m}}{0.80\ \text{kg}}} = 17.3$ rad/s. So $A' = \dfrac{1.5\ \text{m/s}}{17.3\ \text{rad/s}} = 0.0866$ m.
2. The energy lost. $\tfrac12 \times 240\ \text{N/m} \times \left(0.10^2 - 0.0866^2\right)\ \text{m}^2 = 0.30$ J.
3. At the end. The amplitude stays $0.10$ m. The new period is $2\pi\sqrt{\dfrac{0.80\ \text{kg}}{240\ \text{N/m}}} = 0.363$ s. No energy is lost.

At any other point $x$, work out the speed there, share the momentum, then add the potential energy $\tfrac12 kx^2$ again. That gives the new energy and the new amplitude.

Blocks that slip

A block of mass $m$ sits on a larger block of mass $M$. The larger block is joined to a wall by a spring, and it slides on a smooth table. The friction between the two blocks has a coefficient $\mu$.

The top block can only be pulled along by friction. The greatest friction force is $\mu mg$, so its greatest acceleration is $\mu g$. If the pair swings together, the greatest acceleration is $\omega^2A$, with $\omega^2 = \dfrac{k}{M + m}$. The top block holds on while $\omega^2A \le \mu g$, so

A worked example. $M = 2.0$ kg, $m = 1.0$ kg, $k = 300\ \text{N/m}$ and $\mu = 0.40$.

$A_{max} = \dfrac{0.40 \times 9.8\ \text{m/s}^2 \times 3.0\ \text{kg}}{300\ \text{N/m}} = 0.0392$ m, which is $3.92$ cm. At a larger amplitude the top block slides back and forth on the lower one.

Practise with new numbers

Try it in Python

The program finds the effective spring constant of the pulley system from its energy alone. It takes a small step $h$ on each side of the balance point and uses the stored energy there.

Common mistakes

• Putting the new weight in the period. Only $k$ and $m$ set the period. A steady extra acceleration or a different $g$ shifts the middle of the swing and does not change the period.
• Taking the spring stretch as $y$ for a movable pulley. The stretch is $2y$, so the spring constant counts four times.
• Forgetting the starting point. A block that starts at rest at the old middle swings with amplitude $s$ about the new middle, not about the old one.
• Treating the putty at every phase alike. At the end of the swing no energy is lost. At the middle the most is lost.
• Leaving the top block out of $\omega^2$. Both blocks move together, so both masses count in the swinging mass.

Remember these

• Find $k_{eff}$ from the restoring force or the energy for a small move $x$. Then $T = 2\pi\sqrt{\dfrac{m}{k_{eff}}}$.
• A movable pulley that doubles the stretch of the spring gives $k_{eff} = 4k$.
• A support with acceleration $a$ moves the middle by $s = \dfrac{ma}{k}$ and leaves the period alone.
• Putty at the middle gives $A' = A\sqrt{\dfrac{M}{M + m}}$. Putty at the end leaves $A$ unchanged.
• Blocks stay together while $A \le \dfrac{\mu g(M + m)}{k}$.

Check yourself

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

1. A block hangs on a spring in a lift. The lift speeds up upwards. Does the period of the block's swings change? About which point does the block swing now?

2. A block hangs from a movable pulley as in this lesson. What is the ratio of its period to the period of the same block hung straight from the same spring?

3. Putty falls on a swinging block at the end of its swing. Does the amplitude change? Does the period change?

4. A block of $1.0$ kg on a spring of $64\ \text{N/m}$ sits on a trolley. The trolley speeds up at $5.0\ \text{m/s}^2$. How far behind its old middle does the block go at the farthest?

References

• OpenStax, University Physics Volume 1, Chapter 15, Oscillations.
• D. Halliday, R. Resnick and J. Walker, Fundamentals of Physics, the chapter on oscillations.
• I. E. Irodov, Problems in General Physics, the section on mechanical oscillations.