Questions on the mass on a spring and the simple pendulum

Question 1

State the equation for the period of a mass–spring system, defining each symbol.

Accepted Answer

T = 2π√(m/k) (1), where T is the period, m the oscillating mass and k the spring constant (1).

Question 2

Calculate the period of a simple pendulum of length 1.0 m (g = 9.81 m s⁻²).

Accepted Answer

T = 2π√(l/g) = 2π√(1.0/9.81) (1) T = 2.01 s (1)

Question 3

State how the kinetic energy and potential energy of an oscillator vary during one oscillation.

Accepted Answer

They interchange: kinetic energy is greatest at the equilibrium position (where potential energy is least) (1), and potential energy is greatest at the extremes (where kinetic energy is zero) (1). The total energy stays constant.

Question 4

State the condition under which the motion of a simple pendulum is simple harmonic.

Accepted Answer

The oscillations must be of small amplitude (small angle of swing) (1).

Question 5

The mass hanging on a spring is quadrupled. State the effect on the period of oscillation.

Accepted Answer

The period doubles, because T ∝ √m (1).

Question 6

State two quantities that have no effect on the period of a simple pendulum.

Accepted Answer

The mass of the bob (1); and the amplitude, provided the swings are small (1).

Question 7

State what is meant by critical damping, and give one example of a system that is deliberately critically damped.

Accepted Answer

Critical damping is the amount of damping that returns a displaced system to its equilibrium position in the shortest possible time without it oscillating (1). A car suspension, or the needle of a moving-coil meter (1).

Question 8

A 0.50 kg mass on a spring of spring constant 200 N m⁻¹ oscillates with SHM. Calculate the period.

Accepted Answer

T = 2π√(m/k) = 2π√(0.50/200) (1) T = 0.314 s (1)

Question 9

Calculate the length of a simple pendulum that has a period of 2.0 s (g = 9.81 m s⁻²).

Accepted Answer

From T = 2π√(l/g): l = g(T/2π)² (1) = 9.81 × (2.0/2π)² (1) l = 0.994 m (1)

Question 10

A 0.20 kg mass on a spring oscillates with a period of 0.80 s. Calculate (a) the spring constant and (b) the frequency of oscillation.

Accepted Answer

(a) k = 4π²m/T² = 4π² × 0.20/0.80² (1) k = 12.3 N m⁻¹ (1) (b) f = 1/T = 1/0.80 (1) f = 1.25 Hz (1)

Question 11

A mass on a spring (k = 50 N m⁻¹) oscillates with an amplitude of 0.10 m. Calculate the total energy of the oscillation.

Accepted Answer

Total energy = maximum elastic PE = ½kA² (1) = ½ × 50 × 0.10² (1) E = 0.25 J (1)

Question 12

On a mountain expedition, a student sets up a simple pendulum of length 0.800 m and times 20 complete oscillations at 35.9 s. Determine the value of g at the site.

Accepted Answer

T = 35.9/20 = 1.795 s (1) From T = 2π√(l/g): g = 4π²l/T² = 4π² × 0.800/1.795² (1) g = 9.80 m s⁻² (1)

Question 13

A 0.30 kg mass hung from a spring stretches it by 0.045 m (g = 9.81 m s⁻²). Calculate the spring constant. Go on to calculate the period of small vertical oscillations of the mass on this spring.

Accepted Answer

At equilibrium kx = mg, so k = mg/x = (0.30 × 9.81)/0.045 (1) k = 65.4 N m⁻¹ (1) T = 2π√(m/k) = 2π√(0.30/65.4) (1) T = 0.43 s (1)

Question 14

A mass on a spring of spring constant 80 N m⁻¹ oscillates with amplitude 0.060 m. Calculate the total energy of the oscillation and the kinetic energy of the mass when its displacement is 0.030 m.

Accepted Answer

Total energy = ½kA² = ½ × 80 × 0.060² = 0.144 J (1) PE at x = 0.030 m: ½kx² = ½ × 80 × 0.030² = 0.036 J (1) KE = total − PE = 0.144 − 0.036 = 0.108 J (1)

Question 15

An undamped object oscillates with simple harmonic motion. Sketch, on one set of axes, how its kinetic energy and its potential energy vary with time through one complete oscillation, and state how the frequency of each energy variation compares with the frequency of the motion.

Accepted Answer

Both curves run between zero and the same maximum value, and at every instant they add to the same constant total (1). Each is at its maximum where the other is zero, the kinetic energy peaking as the object passes through the centre and the potential energy peaking at the extremes (1). Each energy completes two cycles in one oscillation, so it varies at twice the frequency of the motion, because the two extremes of a cycle give the same energies as each other (1).

Question 16

The bob of a pendulum has a mass of 0.25 kg. It oscillates with a period of 1.6 s and an amplitude of 0.040 m. Calculate the total energy of the oscillation, using E = ½mω²x₀².

Accepted Answer

ω = 2π/T = 2π/1.6 = 3.93 rad s⁻¹ (1) E = ½ × 0.25 × 3.93² × 0.040² (1) E = 3.1 × 10⁻³ J (1)

Question 17

In required practical 7 a student hangs a mass m from a spring, sets it oscillating vertically and times 20 complete oscillations. The results are 0.100 kg, 8.9 s; 0.200 kg, 12.6 s; 0.300 kg, 15.4 s; 0.400 kg, 17.8 s; 0.500 kg, 19.9 s. The student plots T² against m and the points lie on a straight line through the origin. Determine the gradient of that line, and use it to find the spring constant of the spring.

Accepted Answer

T = (time)/20, giving T = 0.445 s at m = 0.100 kg and T = 0.995 s at m = 0.500 kg (1)
Squaring: T² = 0.198 s² and 0.990 s² (1)
Gradient = ΔT²/Δm = (0.990 − 0.198)/(0.500 − 0.100) = 1.98 s² kg⁻¹ (accept 1.9 to 2.1) (1)
Squaring T = 2π√(m/k) gives T² = (4π²/k)m, so the gradient is 4π²/k and k = 4π²/gradient (1)
k = 4π²/1.98 = 19.9 N m⁻¹ (accept 19 to 21) (1)

Question 18

A simple pendulum of length 1.0 m is taken to the Moon, where g = 1.62 m s⁻². Calculate its period there and compare it with its period on Earth (2.0 s).

Accepted Answer

T = 2π√(l/g) = 2π√(1.0/1.62) (1) T = 4.94 s (1) This is much longer than on Earth (1), because the weaker gravity gives a smaller restoring effect (1)

Question 19

A 0.30 kg mass on a spring of spring constant 120 N m⁻¹ oscillates with an amplitude of 0.050 m. Calculate its maximum speed.

Accepted Answer

ω = √(k/m) = √(120/0.30) = 20 rad s⁻¹ (1) v_max = ωA = 20 × 0.050 (1) v_max = 1.00 m s⁻¹ (1)

Question 20

Describe the effect of damping on an oscillation.

Accepted Answer

Damping (resistive forces such as friction or air resistance) removes energy from the oscillator (1), so the amplitude decreases with each cycle until the motion stops (1). Light damping barely changes the period; heavy damping can stop the system oscillating at all (1).

Question 21

A grandfather clock keeps correct time using a simple pendulum of length 0.994 m where g = 9.81 m s⁻²; its period there is 2.000 s. The clock is moved to a city where g = 9.78 m s⁻². Deduce whether the clock now runs fast or slow, determine the time it gains or loses per day, and suggest how the pendulum should be adjusted to correct it.

Accepted Answer

New period T = 2π√(0.994/9.78) (1)
T = 2.0031 s, longer than 2.000 s (1)
Each swing takes longer, so the clock counts too few swings per day: it runs slow (1)
Time lost per day = 86 400 × (2.00310 − 2.00004)/2.00310 ≈ 132 s (accept 125 to 135 s) (1)
Shorten the pendulum (so that 2π√(l/9.78) = 2.000 s again) (1)

Question 22

A student sets a mass-spring system oscillating vertically in air, then repeats the experiment with a large card attached to the mass to increase air resistance. Describe and explain how the displacement-time graph and the energy of the system differ between the two experiments.

Accepted Answer

In both cases the graph is an oscillation whose amplitude decreases with time, since air resistance damps the motion (1). With the card the damping force is larger, so the amplitude decays much more quickly (1). The decay is roughly exponential: the amplitude falls by a similar fraction each cycle (1). The period is almost unchanged by this light damping, so the two graphs keep nearly the same spacing of peaks (1). The total mechanical energy of the oscillator decreases as work is done against the resistive force, and it falls faster with the card attached (1). The energy is transferred to the surroundings as internal energy (of the air and the system) (1).

Question 23

Liquid in a vertical U-tube oscillates with a period T = 2π√(L/2g), where L is the total length of the liquid column. Calculate the period when L = 0.30 m (g = 9.81 m s⁻²), and state and explain what happens to the period if a denser liquid is used with the same column length.

Accepted Answer

T = 2π√(0.30/(2 × 9.81)) (1) T = 0.78 s (1) No change: density does not appear in the formula, because both the restoring force and the mass of the column scale with density, so it cancels (1)

Question 24

A mass on a spring of spring constant 60 N m⁻¹ oscillates with an amplitude of 0.080 m. Calculate the total energy of the oscillation, and determine the displacement from equilibrium at which the kinetic energy of the mass is equal to its potential energy.

Accepted Answer

Total energy = ½kA² = ½ × 60 × 0.080² = 0.192 J (1) The two are equal when each is half the total, 0.096 J (1) ½kx² = 0.096, so x = √((2 × 0.096)/60) (1) x = 0.057 m, which is A/√2 (1)

Question 25

In required practical 7 a student times 20 complete oscillations of a simple pendulum at each of five lengths, recording the results in the table. The student plots T² on the vertical axis against l on the horizontal axis and the points lie on a straight line through the origin. Determine the gradient of that line, and use it to obtain a value for the gravitational field strength g at the laboratory.

Accepted Answer

T = t/20, giving T = 25.37/20 = 1.269 s at l = 0.400 m and T = 43.95/20 = 2.198 s at l = 1.200 m (1)
Squaring those periods gives T² = 1.61 s² and T² = 4.83 s² (1)
Gradient = (4.83 − 1.61)/(1.200 − 0.400) = 3.22/0.800 = 4.025 s² m⁻¹ (1)
Comparing with T² = (4π²/g)l, the gradient is 4π²/g, so g = 4π²/gradient (1)
g = 39.5/4.025 = 9.81 m s⁻² (accept 9.7 to 9.9 m s⁻²) (1)

Question 26

A mass hangs from a spring of spring constant 25 N m⁻¹ and is set oscillating vertically in air. The amplitude at each successive maximum displacement is recorded in the table. Determine the total energy of the oscillation at t = 0 and at t = 6.0 s, and deduce which grade of damping the readings show, justifying your answer from the numbers.

Accepted Answer

E = ½kA², with the amplitude converted to metres, A = 60.0 mm = 0.0600 m (1)
At t = 0, E = ½ × 25 × 0.0600² = 4.5 × 10⁻² J (1)
At t = 6.0 s, A = 24.6 mm and E = ½ × 25 × 0.0246² = 7.6 × 10⁻³ J, about 17% of the original energy, since E goes as A² (1)
The damping is light: the system goes on oscillating, and the maxima stay a constant 1.5 s apart, so the period is unchanged as the amplitude falls (1)
Each amplitude is 0.80 of the one before (48.0/60.0 = 38.4/48.0 = 0.80), the constant ratio of an exponential envelope; heavy or critical damping would have permitted no oscillation at all (1)

length l / m	time for 20 oscillations t / s
0.400	25.37
0.600	31.08
0.800	35.89
1.000	40.12
1.200	43.95