Worked examples · Sheet 2

Mass-spring oscillators

Theory used: Mass-spring oscillator Vertical spring Springs in parallel and in series Velocity and acceleration in harmonic motion Amplitude and phase from the initial conditions

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A body of mass $m = 0.50 kg$ , resting on a smooth horizontal plane, is attached to an ideal spring with spring constant $k = 200 N/m$ . The body is displaced by $4.0 cm$ from the rest position and released from rest. Determine the angular frequency, the period, the maximum velocity, the maximum acceleration and the maximum magnitude of the elastic force.

m \overset{x}{¨} = - k x ⟹ ω = \frac{k}{m}, T = 2 π \frac{m}{k}

ω = \frac{200}{0.50} rad/s = 20 rad/s, T = \frac{2 π}{20} s = 0.314 s

A = 0.040 m, v_{m a x} = A ω = 0.80 m/s, a_{m a x} = A ω^{2} = 16 m/ s^{2}

F_{m a x} = k A = 200 \cdot 0.040 N = 8.0 N

The plane is smooth and the motion is horizontal, so the weight and the normal force balance and along the axis of the spring only the elastic force $- k x$ acts. Newton’s second law gives $m \overset{x}{¨} = - k x$ , which is the equation of harmonic motion with $ω^{2} = \frac{k}{m}$ ; therefore the body is a mass-spring oscillator. The body starts from rest $4.0 cm$ from the rest position, so this displacement is the amplitude. The maximum velocity and acceleration follow from the formulas of harmonic motion. The elastic force is maximum at the extremes, where $∣ x ∣ = A$ ; as a check, $\frac{F _{m a x}}{m} = \frac{8.0}{0.50} m/ s^{2} = 16 m/ s^{2}$ , which coincides with $a_{m a x}$ .

Exercise 2

A body of mass $m = 0.30 kg$ hanging from an ideal vertical spring stretches it, at equilibrium, by $Δ ℓ = 6.0 cm$ . Determine the spring constant and the period of the vertical oscillations. The body is then pulled down a further $2.0 cm$ and released from rest; determine the maximum velocity.

k Δ ℓ = m g ⟹ k = \frac{m g}{Δ ℓ} = \frac{0.30 \cdot 9.81}{0.060} N/m = 49.1 N/m

ω = \frac{k}{m} = \frac{g}{Δ ℓ} = \frac{9.81}{0.060} rad/s = 12.8 rad/s

T = \frac{2 π}{ω} = 2 π \frac{Δ ℓ}{g} = 0.491 s

v_{m a x} = A ω = 0.020 \cdot 12.8 m/s = 0.256 m/s

On the left the body at equilibrium, with the spring stretched by

Δ ℓ

with respect to its rest length; on the right the forces on the body: the elastic force

k Δ ℓ

, pointing upward, balances the weight

m g

At equilibrium the acceleration is zero, so the elastic force, which equals $k Δ ℓ$ and points upward, balances the weight; from this condition we obtain $k$ . By the theorem on the vertical spring the body oscillates about the equilibrium position with the same angular frequency as the horizontal spring, and since $\frac{k}{m} = \frac{g}{Δ ℓ}$ the period is computed from the static extension alone, without using the mass. The body is released from rest $2.0 cm$ below the equilibrium position, which is the center of the oscillation, so the amplitude is $2.0 cm$ , and the maximum velocity is $A ω$ . The total extension of $8.0 cm$ is not the amplitude, because it is measured from the rest length.

Exercise 3

Two ideal springs with constants $k_{1} = 300 N/m$ and $k_{2} = 600 N/m$ hold a body of mass $m = 2.0 kg$ that oscillates on a smooth horizontal plane. Compute the period when the springs are connected in parallel and when they are connected in series, and the ratio of the two periods.

k_{p} = k_{1} + k_{2} = 900 N/m, k_{s} = \frac{k _{1} k _{2}}{k _{1} + k _{2}} = \frac{300 \cdot 600}{900} N/m = 200 N/m

T_{p} = 2 π \frac{m}{k _{p}} = 2 π \frac{2.0}{900} s = 0.296 s

T_{s} = 2 π \frac{m}{k _{s}} = 2 π \frac{2.0}{200} s = 0.628 s

\frac{T _{s}}{T _{p}} = \frac{k _{p}}{k _{s}} = \frac{900}{200} = 2.12

Two springs in parallel have the same extension and their forces add up, so they are equivalent to a spring with constant $k_{1} + k_{2}$ . Two springs in series carry the same force and their extensions add up, so they are equivalent to a spring with constant $\frac{k _{1} k _{2}}{k _{1} + k _{2}}$ , smaller than both. With the equivalent constant the system is a mass-spring oscillator, therefore the period is $2 π \frac{m}{k}$ . The period is inversely proportional to the square root of $k$ , so the series connection, which is softer, gives the longer period, about $2.1$ times that in parallel.

Exercise 4

A body of unknown mass attached to an ideal spring oscillates on a smooth horizontal plane with period $T_{1} = 0.60 s$ . When a mass $Δ m = 0.20 kg$ is added to the body, the period becomes $T_{2} = 0.75 s$ . Determine the mass of the body and the spring constant.

T_{1}^{2} = 4 π^{2} \frac{m}{k}, T_{2}^{2} = 4 π^{2} \frac{m + Δ m}{k} ⟹ \frac{T _{2}^{2}}{T _{1}^{2}} = \frac{m + Δ m}{m}

m = \frac{Δ m}{\frac{T _{2}^{2}}{T _{1}^{2}} - 1} = \frac{0.20}{( \frac{0.75}{0.60} ) ^{2} - 1} kg = \frac{0.20}{0.5625} kg = 0.356 kg

k = \frac{4 π ^{2} m}{T _{1}^{2}} = \frac{4 π ^{2} \cdot 0.356}{0.6 0 ^{2}} N/m = 39.0 N/m

The period of the mass-spring oscillator is $2 π \frac{m}{k}$ , so its square is proportional to the mass. The spring is the same in the two measurements, therefore dividing the two relations the constant $k$ cancels and an equation in the mass $m$ alone remains. Once the mass is known, the spring constant is obtained from the first relation. As a check, with the total mass $0.556 kg$ we obtain $2 π \frac{0.556}{39.0} s = 0.75 s$ , which is the second measured period. The method does not require the weight of the body, so it also works in the absence of gravity.