Unit 7: Oscillations

Simple Harmonic Motion

Oscillations will mostly deal with one specific type of motion: Simple Harmonic Motion (SHM).

Definition (Simple Harmonic Motion). A system is in simple harmonic motion when its acceleration is proportional to displacement and points toward equilibrium, $a = -\omega^2 x$ , equivalently $\dfrac{d^2x}{dt^2} + \omega^2x = 0$ .

The general solution is given as:

x(t)=A\cos(\omega t+\phi)

or equivalently

x(t)=A\sin(\omega t+\phi).

A rigorous proof of the formula (one that does not involve knowing the solution beforehand) requires multivariable calculus and will not be shown here. The constants $A$ and $\phi$ are set by initial position and velocity. Concretely, from $x(t)=A\cos(\omega t+\phi)$ and $v(t)=-A\omega\sin(\omega t+\phi)$ , evaluating at $t=0$ gives

x_0=A\cos\phi,\qquad v_0=-A\omega\sin\phi.

Solving these two equations for the two unknowns yields

A=\sqrt{x_0^2+\left(\frac{v_0}{\omega}\right)^2},\qquad \tan\phi=-\frac{v_0}{\omega x_0}.

There are many types of simple harmonic motion (the most common of which are the spring and the simple pendulum), and most problems will state that it involves SHM.

Velocity and Acceleration in SHM

x(t)=A\cos(\omega t+\phi),

then taking the derivative gets

v(t)=\frac{dx}{dt}=-A\omega\sin(\omega t+\phi),

and taking it again gets

a(t)=\frac{d^2x}{dt^2}=-A\omega^2\cos(\omega t+\phi)=-\omega^2x.

The maximum speed is

v_{\max}=A\omega,

at a point where $\sin x = 1$ and the maximum acceleration magnitude is

a_{\max}=A\omega^2.

Speed is greatest at equilibrium and zero at the turning points, both of which are pretty intuitive.

Example. A mass in SHM has amplitude $A=0.20\ \text{m}$ . At $x=0.12\ \text{m}$ , its speed is $0.80\ \text{m/s}$ . Find its maximum speed, without using the period or frequency. For SHM, it may be useful to know that $v^2=\omega^2(A^2-x^2).$

For SHM,

v^2=\omega^2(A^2-x^2)

and

v_{\max}=A\omega.

Eliminate $\omega$ :

\frac{v}{v_{\max}}=\frac{\sqrt{A^2-x^2}}{A}.

v_{\max}=v\frac{A}{\sqrt{A^2-x^2}} =0.80\frac{0.20}{\sqrt{0.20^2-0.12^2}}=1.0\ \text{m/s}.

Period and Frequency

The angular frequency $\omega$ is related to frequency and period by

\omega = 2\pi f = \frac{2\pi}{T}.

Thus

T = \frac{2\pi}{\omega}.

Period is the time for one full cycle. Frequency is cycles per second, measured in hertz.

Angular frequency is not “cycles per second”; it is radians of phase per second. One cycle corresponds to $2\pi$ radians of phase, which is why $\omega=2\pi f$ . In SHM, $\omega$ is usually the most natural quantity because it appears directly in $a=-\omega^2x$ . Large $\omega$ means a strong restoring acceleration for a given displacement, so the oscillator turns around quickly and has a short period.

Example. An oscillator has position $x(t)=0.12\cos(8.0t+\pi/3)$ in meters. Find its amplitude, angular frequency, period, and frequency.

Compare the equation to $x(t)=A\cos(\omega t+\phi)$ :

A=0.12\ \text{m},\qquad \omega=8.0\ \text{rad/s}.

The period is

T=\frac{2\pi}{\omega} =\frac{2\pi}{8.0} =0.785\ \text{s}.

The frequency is

f=\frac{1}{T} =\frac{\omega}{2\pi} =\frac{8.0}{2\pi} =1.27\ \text{Hz}.

The phase $\pi/3$ shifts where the oscillator starts, but it does not change amplitude, period, or frequency.

Common Modes of SHM

Mass-Spring Oscillators

For a mass on an ideal, massless spring, the force follows Hooke’s Law:

F=-kx.

The angular frequency is defined as $\omega=\sqrt{\frac{k}{m}}$ and the period is $T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{k}}.$ As a reminder, $k$ is the spring constant, and has units of $\frac{N}{m}$ . A larger spring constant means the spring is more stiff and thus takes more force to push or pull on it.

Proof (Mass-Spring Period). Hooke’s law gives

F=ma=-kx.

Using the differential form of Newton’s second law gives

m\frac{d^2x}{dt^2}=-kx.

Rearrange:

\frac{d^2x}{dt^2}+\frac{k}{m}x=0.

Compare this to the SHM form (since all springs undergo SHM)

\frac{d^2x}{dt^2}+\omega^2x=0.

\omega^2=\frac{k}{m},

meaning

\omega=\sqrt{\frac{k}{m}}.

Since $T=2\pi/\omega$ ,

T=2\pi\sqrt{\frac{m}{k}}.

Combinations of springs

Many AP setups attach a mass to more than one spring. Each combination behaves like a single ideal spring with an effective spring constant $k_{\text{eff}}$ that models the whole system, and the period is then $T=2\pi\sqrt{m/k_{\text{eff}}}$ . Similarly, $U=\frac{1}{2}k_{\text{eff}}x^2$ .

Springs are in parallel when they attach to the same moving object and stretch or compress by the same amount. If the mass moves by $x$ , each parallel spring changes length by $x$ , so their restoring forces add.

Springs are in series when they are connected end-to-end and the same force passes through each spring. The total stretch is split between them: one spring may stretch more than the other, but both carry the same tension. A good test is: same displacement means parallel; same force through each spring means series.

Proof (series and parallel spring constants). Lets say you have two ideal springs with spring constant $k_1$ and $k_2$ supporting one mass.

Parallel. Each pulls back, and the forces add:

F=-k_1 x-k_2 x=-(k_1+k_2)x.

Comparing to $F=-k_{\text{eff}}x$ ,

k_{\text{eff}}=k_1+k_2.

Parallel springs are stiffer than either alone. If you have more than two springs in parallel, you simply add all of the spring constants together to get the effective spring constant.

Series. Each spring stretches by $x_i=F/k_i$ in magnitude, and the total stretch is

x=x_1+x_2=\frac{F}{k_1}+\frac{F}{k_2}=F\left(\frac{1}{k_1}+\frac{1}{k_2}\right).

Since $k_{\text{eff}}=F/x$ ,

\frac{1}{k_{\text{eff}}}=\frac{1}{k_1}+\frac{1}{k_2}.

Alternatively,

k_{\text{eff}}=\frac{k_1 k_2}{k_1 + k_2}

Series springs are softer than either alone. If you have more than two springs in parallel, you simply add all of the reciprocals of the spring constants together and take the reciprocal of the sum to get the effective spring constant.

Example. A $2.0\ \text{kg}$ block is connected to two springs with $k_1=300\ \text{N/m}$ and $k_2=600\ \text{N/m}$ . Find the period (a) if the springs act in parallel and (b) if they act in series.

(a) Parallel:

k_{\text{eff}}=k_1+k_2=300+600=900\ \text{N/m},

T=2\pi\sqrt{\frac{m}{k_{\text{eff}}}}=2\pi\sqrt{\frac{2.0}{900}}\approx 0.30\ \text{s}.

(b) Series:

\frac{1}{k_{\text{eff}}}=\frac{1}{300}+\frac{1}{600}=\frac{2+1}{600}=\frac{1}{200}\quad\Longrightarrow\quad k_{\text{eff}}=200\ \text{N/m},

T=2\pi\sqrt{\frac{2.0}{200}}\approx 0.63\ \text{s}.

The series arrangement is softer, so it oscillates more slowly. A quick sanity check: $200\ \text{N/m}$ is smaller than either individual spring, and $900\ \text{N/m}$ is larger than either — exactly as the proof predicts.

Effective spring constants from energy

Not every spring configuration is purely series or parallel. A general method is to displace the mass by a small coordinate $x$ (which simulates SHM), write the total spring potential energy, and match it to

U=\frac{1}{2}k_{\text{eff}}x^2.

Example. A block is attached symmetrically to two identical springs of constant $k$ . Each spring makes angle $\theta$ with the horizontal at equilibrium. If the block is displaced a small distance $x$ horizontally, find the effective spring constant for horizontal oscillations.

For a small horizontal displacement, each spring’s length changes by the component of the block’s displacement along that spring:

\Delta \ell=x\cos\theta.

Each spring stores energy $\tfrac12k(\Delta\ell)^2$ , so the total spring energy is

U=2\left(\frac12k(x\cos\theta)^2\right)=kx^2\cos^2\theta.

Match this to $\tfrac12k_{\text{eff}}x^2$ :

\frac12k_{\text{eff}}x^2=kx^2\cos^2\theta.

Therefore

k_{\text{eff}}=2k\cos^2\theta.

This configuration is not simply series or parallel; the geometry determines how much each spring actually stretches.

Simple Pendulums

A simple pendulum is defined as a point mass (or equivalent, explained in the next section) hanging on a massless rod that swings at small angles.

Definition (Simple Pendulum). For a simple pendulum of length $L$ and small angular displacement $\theta$ ,

\omega=\sqrt{\frac{g}{L}},

and

T=2\pi\sqrt{\frac{L}{g}}.

This formula assumes the small-angle approximation $\sin\theta\approx\theta$ when $\theta$ is measured in radians. Like the ideal mass-spring period, the simple-pendulum period is independent of mass and approximately independent of amplitude.

Proof (Simple Pendulum Period). Gravity creates a restoring torque about the pivot:

\tau=-mgL\sin\theta.

For small angles, $\sin\theta\approx\theta$ (using a Taylor series expansion), so

\tau\approx-mgL\theta.

Using $I=mL^2$ and $\sum\tau=I\,d^2\theta/dt^2$ ,

mL^2\frac{d^2\theta}{dt^2}=-mgL\theta.

Rearrange:

\frac{d^2\theta}{dt^2}+\frac{g}{L}\theta=0.

Compare with $d^2\theta/dt^2+\omega^2\theta=0$ :

\omega=\sqrt{\frac{g}{L}},

T=2\pi\sqrt{\frac{L}{g}}.

Note that this only works for small angles of $\theta$ ! Without the small angle approximation, the solution has no closed form and requires elliptical integrals to solve!

Example. Suppose a pendulum has a length of $1$ meter and a period of $2$ seconds. Find $g$ in this scenario (do NOT assume that $g=9.8$ !).

Use

T=2\pi\sqrt{\frac{L}{g}}.

Solve for $g$ :

\frac{T}{2\pi}=\sqrt{\frac{L}{g}},

\frac{T^2}{4\pi^2}=\frac{L}{g},

g=\frac{4\pi^2L}{T^2}.

With $L=1\ \text{m}$ and $T=2\ \text{s}$ ,

g=\frac{4\pi^2(1)}{2^2}=\pi^2\ \text{m/s}^2.

Approximately,

g\approx9.87\ \text{m/s}^2.

You may find it curious that the solution ( $\pi^2$ ) is very close to $g$ , the gravitational acceleration on Earth! This is no coincidence. In the late 18th century, scientists defined the meter as the length of a “seconds pendulum,” where the time it takes for it to go from end to end is one second (and thus the period is two seconds).

However, as time went on, scientists changed the definition of the meter since gravity was different depending on where you stand from, resulting in a geodetic definition: one ten-millionth of the distance from the equator to the North Pole along the meridian passing through Paris. The old definition ended up being around 1% off of the current accepted value for $g$ !

Physical Pendulums

While simple pendulums are very easy to solve, not all pendulums act like point masses. A rigid body swinging about a pivot is known as a physical pendulum and is subject to rotation about the pivot point. If the center of mass is distance $d$ from the pivot,

\omega=\sqrt{\frac{mgd}{I}},

and

T=2\pi\sqrt{\frac{I}{mgd}}.

Proof (Physical Pendulum Period). Gravity acts at the center of mass, distance $d$ from the pivot. The torque is

\tau=-mgd\sin\theta.

For small angles, $\sin\theta\approx\theta$ , so

\tau\approx-mgd\theta.

Using $\sum\tau=I\,d^2\theta/dt^2$ about the pivot,

I\frac{d^2\theta}{dt^2}=-mgd\theta.

Rearrange:

\frac{d^2\theta}{dt^2}+\frac{mgd}{I}\theta=0.

Thus

\omega=\sqrt{\frac{mgd}{I}},

and

T=2\pi\sqrt{\frac{I}{mgd}}.

Here $I$ is the rotational inertia about the pivot, not about the center of mass; use the parallel-axis theorem when needed. The simple pendulum is the special case where all mass sits at distance $L$ from the pivot (for all purposes, this is a point mass), giving $I=mL^2$ and $d=L$ , which reduces $T$ back to $2\pi\sqrt{L/g}$ .

Example. A uniform rod of mass $M$ and length $L=1.2\ \text{m}$ is pivoted at one end and swings as a physical pendulum through small angles. Find its period. Use $g=9.8\ \text{m/s}^2$ .

The rotational inertia of a uniform rod about one end is

I=\frac{1}{3}ML^2.

The center of mass is at the rod’s midpoint, so $d=L/2$ . Substitute into the physical-pendulum period:

T=2\pi\sqrt{\frac{I}{Mgd}}=2\pi\sqrt{\frac{\tfrac{1}{3}ML^2}{Mg\,(L/2)}}=2\pi\sqrt{\frac{2L}{3g}}.

The mass $M$ cancels, as it must for a gravity-driven pendulum. Plugging in numbers:

T=2\pi\sqrt{\frac{2(1.2)}{3(9.8)}}=2\pi\sqrt{\frac{2.4}{29.4}}=2\pi\sqrt{0.0816}\approx 1.8\ \text{s}.

It is instructive to compare this to a simple pendulum of the same length $L=1.2\ \text{m}$ , which would have $T=2\pi\sqrt{L/g}=2\pi\sqrt{1.2/9.8}\approx2.2\ \text{s}$ . The rod swings faster because its mass is distributed closer to the pivot than a point mass at the far end — equivalently, the rod behaves like a simple pendulum of effective length $\tfrac{2}{3}L=0.80\ \text{m}$ .

Solving for general SHM systems

Not every oscillator is a spring or a pendulum, but the same methods still apply.

Example. A solid cylinder of cross-sectional area $A$ , height $h$ , and density $\rho_{\text{obj}}$ floats upright in a liquid of density $\rho_{\text{liq}}$ . It is pushed down slightly and released. Show the motion is SHM and find $\omega$ .

These formulas may be helpful:

F_B=\rho_{\text{fluid}}gV_{\text{sub}}, \qquad m=\rho_{\text{object}}V.

The first is Archimedes’ principle: buoyant force equals the weight of displaced fluid. The second is the density relation, mass equals density times volume.

At equilibrium the object floats with some submerged depth $d_0$ , where buoyancy balances weight. Now push it down an extra distance $y$ . The submerged volume increases by $A\,y$ , so the buoyant force grows by

\Delta F_B=\rho_{\text{liq}}\,g\,(A\,y),

directed upward — opposite the displacement. Since the weight is unchanged, the net restoring force is

F=-\rho_{\text{liq}}\,g\,A\,y.

This is Hooke’s law with effective constant $k_{\text{eff}}=\rho_{\text{liq}}gA$ . The object’s mass is $m=\rho_{\text{obj}}Ah$ , so

\omega=\sqrt{\frac{k_{\text{eff}}}{m}}=\sqrt{\frac{\rho_{\text{liq}}gA}{\rho_{\text{obj}}Ah}}=\sqrt{\frac{\rho_{\text{liq}}\,g}{\rho_{\text{obj}}\,h}}.

The area $A$ cancels, and the frequency depends only on the density ratio, $g$ , and the height — a clean result that comes entirely from the linear restoring force.

Example. A U-shaped tube of uniform cross-sectional area $A$ contains a liquid of density $\rho$ , with total liquid column length $L$ . The liquid is disturbed so one side rises by $y$ while the other falls by $y$ . Find the period of the resulting oscillation.

For a uniform tube, the liquid volume is cross-sectional area times length, $V=AL$ (since the object is a cylinder). Weight is then the gravitational force on that mass.

When the left surface drops by $y$ and the right rises by $y$ , the height difference between the two columns is $2y$ . That excess column of height $2y$ and cross-section $A$ has weight

\Delta W=\rho g\,(A\cdot 2y),

and this unbalanced weight is the restoring force on the whole liquid:

F=-2\rho g A\,y.

The total moving mass is the entire liquid column, $m=\rho A L$ . Newton’s second law gives

\rho A L\frac{d^2y}{dt^2}=-2\rho g A\,y \quad\Longrightarrow\quad \frac{d^2y}{dt^2}+\frac{2g}{L}\,y=0.

\omega=\sqrt{\frac{2g}{L}},\qquad T=2\pi\sqrt{\frac{L}{2g}}.

The density and area both cancel: only the column length matters. As with the floating object, the entire problem reduced to writing the restoring force as a constant times the displacement.

Energy in SHM

For a mass-spring oscillator,

E = K+U = \frac{1}{2}mv^2+\frac{1}{2}kx^2.

At maximum displacement, $v=0$ and $x=\pm A$ , so

E=\frac{1}{2}kA^2.

At equilibrium, $x=0$ and speed is maximum:

E=\frac{1}{2}mv_{\max}^2.

Energy continuously transfers between kinetic energy and spring potential energy while total mechanical energy remains constant if the spring system is ideal. These equations lead to the following two equations:

\frac{1}{2}mv^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2 \quad\Longrightarrow\quad v(x)=\pm\omega\sqrt{A^2-x^2}.

This recovers $v_{\max}=A\omega$ at $x=0$ and $v=0$ at $x=\pm A$ without ever solving the differential equation.

Proof (SHM from energy conservation). For an ideal mass-spring system the total mechanical energy is constant:

E=\frac{1}{2}mv^2+\frac{1}{2}kx^2=\text{const.}

Differentiate both sides with respect to time. Using $v=dx/dt$ and $dv/dt=d^2x/dt^2$ ,

\frac{dE}{dt}=mv\frac{dv}{dt}+kx\frac{dx}{dt}=mv\frac{d^2x}{dt^2}+kxv=0.

Factor out $v$ :

v\left(m\frac{d^2x}{dt^2}+kx\right)=0.

Since $v$ is not zero throughout the motion (it is instantaneously zero only at the turning points), the bracket must vanish:

m\frac{d^2x}{dt^2}+kx=0,

which is exactly the SHM equation with $\omega=\sqrt{k/m}$ . This equation can apply to any type of SHM, not just mass-spring oscillators.

Example. A block oscillates on a spring with amplitude $A=0.12\ \text{m}$ . Find the displacement where the kinetic energy is three times the potential energy, and find the fraction of the maximum speed at that point.

Total energy is $E=\tfrac{1}{2}kA^2$ . When $K=3U$ ,

E=K+U=4U.

Thus $U=E/4$ :

\frac12kx^2=\frac14\left(\frac12kA^2\right).

Cancel $\tfrac12k$ :

x^2=\frac{A^2}{4}\quad\Longrightarrow\quad x=\pm\frac{A}{2}=\pm0.060\ \text{m}.

Since $K=3E/4$ and $K_{\max}=E=\tfrac12mv_{\max}^2$ ,

\frac12mv^2=\frac34\left(\frac12mv_{\max}^2\right).

Therefore

v=\frac{\sqrt3}{2}v_{\max}\approx0.866v_{\max}.

Small Oscillations About Any Potential Minimum

The mass-spring system is just the simplest member of a much larger family. Recall that a one-dimensional conservative force is $F_x=-dU/dx$ and that a stable equilibrium sits at a local minimum of $U(x)$ , where $U'(x_0)=0$ and $U''(x_0)>0$ . The key result is that any such minimum looks like a spring for small displacements, with an effective spring constant equal to the curvature of $U$ (think of a ball rolling back and forth in a divot):

k_{\text{eff}}=U''(x_0),\qquad \omega=\sqrt{\frac{U''(x_0)}{m}}.

Proof (small oscillations are SHM). Let a particle of mass $m$ move in a potential $U(x)$ with a stable equilibrium at $x_0$ , so $U'(x_0)=0$ and $U''(x_0)>0$ . Expand $U$ in a Taylor series (refer to Calc BC Unit 10 if you need more information about this) about $x_0$ , writing the small displacement $s=x-x_0$ :

U(x)=U(x_0)+U'(x_0)\,s+\frac{1}{2}U''(x_0)\,s^2+\cdots

The constant $U(x_0)$ does not affect forces, and $U'(x_0)=0$ by the equilibrium condition. For small $s$ we keep only the quadratic term, since all the other terms are essentially $0$ :

U(x)\approx U(x_0)+\frac{1}{2}U''(x_0)\,s^2.

The force is then

F=-\frac{dU}{dx}=-\frac{dU}{ds}=-U''(x_0)\,s.

This is exactly Hooke’s law with $k_{\text{eff}}=U''(x_0)$ . Newton’s second law gives

m\frac{d^2s}{dt^2}=-U''(x_0)\,s \quad\Longrightarrow\quad \omega=\sqrt{\frac{U''(x_0)}{m}}.

So near the bottom of any smooth potential well, the motion is simple harmonic — the parabola $\tfrac12 k_{\text{eff}}s^2$ is the local approximation to every potential minimum. Another way to think of it is

\omega=\sqrt{\frac{\text{restoring effect}}{\text{inertia}}}.

Here the restoring effect is the curvature $U''(x_0)$ : a sharper well pushes back harder for the same displacement. The inertia is the mass $m$ : a heavier object responds more sluggishly to the same restoring force.

Example. A particle of mass $m=0.20\ \text{kg}$ moves in the potential

U(x)=\alpha x^4-\beta x^2,\qquad \alpha=2.0\ \text{J/m}^4,\quad \beta=4.0\ \text{J/m}^2.

Find the angular frequency of small oscillations about the positive- $x$ equilibrium.

First locate the equilibrium from $U'(x)=0$ :

U'(x)=4\alpha x^3-2\beta x=2x(2\alpha x^2-\beta)=0.

The nonzero solutions are $x_0=\pm\sqrt{\beta/2\alpha}=\pm\sqrt{4.0/4.0}=\pm1.0\ \text{m}$ . Take $x_0=+1.0\ \text{m}$ . Now the curvature:

U''(x)=12\alpha x^2-2\beta,

U''(x_0)=12(2.0)(1.0)^2-2(4.0)=24-8=16\ \text{N/m}.

Since $U''(x_0)>0$ this is indeed a stable minimum (the point $x=0$ , where $U''(0)=-8<0$ , is an unstable maximum). The angular frequency is

\omega=\sqrt{\frac{U''(x_0)}{m}}=\sqrt{\frac{16}{0.20}}=\sqrt{80}\approx 8.9\ \text{rad/s}.

Notice we never needed the full shape of $U$ — only its second derivative at the equilibrium.

Key equations

Idea	Equation
SHM equation of motion	$\dfrac{d^2x}{dt^2}+\omega^2 x=0$
General solution	$x(t)=A\cos(\omega t+\phi)$
Amplitude from initial conditions	$A=\sqrt{x_0^2+(v_0/\omega)^2}$
Max speed / acceleration	$v_{\max}=A\omega,\quad a_{\max}=A\omega^2$
Speed vs position	$v(x)=\pm\omega\sqrt{A^2-x^2}$
Period and frequency	$\omega=2\pi f=2\pi/T$
Mass-spring	$\omega=\sqrt{k/m},\quad T=2\pi\sqrt{m/k}$
Springs in parallel	$k_{\text{eff}}=k_1+k_2$
Springs in series	$1/k_{\text{eff}}=1/k_1+1/k_2$
General spring configuration	match $U$ to $\tfrac12k_{\text{eff}}x^2$
Simple pendulum	$T=2\pi\sqrt{L/g}$
Physical pendulum	$T=2\pi\sqrt{I/(mgd)}$
Energy in SHM	$E=\tfrac12 kA^2=\tfrac12 m v_{\max}^2$
Small oscillations near $U$ minimum	$\omega=\sqrt{U''(x_0)/m}$

Practice

Multiple Choice

A mass $m$ hangs from a spring of constant $k$ in a uniform gravitational field. It oscillates vertically about equilibrium with amplitude $A$ . Which quantity depends on $g$ ?

(A) The angular frequency

(B) The period

(D) The speed at the equilibrium point measured relative to the oscillation amplitude

A pendulum clock is taken to a planet where the gravitational field strength is $g/4$ . To keep the same small-angle period, the pendulum length should be changed from $L$ to

(A) $4L$

(B) $2L$

(D) $L/4$

A mass on a spring is released from rest at $x=A$ . When it first reaches $x=A/3$ , what fraction of the total mechanical energy is kinetic?

(A) $1/9$

(B) $2/3$

(D) $\sqrt{8}/3$

A block of mass $m$ is attached to a spring of constant $k$ on a frictionless horizontal surface. A small constant horizontal force $F_0$ is then applied and left on. Compared with the original oscillator, the new motion has

(A) the same angular frequency and an equilibrium shifted by $F_0/k$

(B) angular frequency $\sqrt{(k+F_0)/m}$ and the same equilibrium

(D) no simple harmonic motion because the net force is not proportional to $x$

A bead slides without friction on a circular hoop of radius $R$ in a vertical plane. Near the bottom of the hoop, the coordinate along the arc is $s=R\theta$ . The bead’s small-oscillation angular frequency is

(A) $\sqrt{g/R}$

(B) $\sqrt{R/g}$

(D) $\sqrt{2g/R}$

A mass $m$ is attached between two horizontal springs with constants $k_1$ and $k_2$ on a frictionless track, one spring on each side. Both springs are relaxed when the mass is at $x=0$ . If the mass is displaced slightly, its angular frequency is

(A) $\sqrt{\dfrac{k_1+k_2}{m}}$

(B) $\sqrt{\dfrac{k_1k_2}{m(k_1+k_2)}}$

(D) $\sqrt{\dfrac{k_1}{m}}+\sqrt{\dfrac{k_2}{m}}$

A block attached to a spring oscillates on a frictionless table. The block is replaced by two identical blocks glued together, and the amplitude is doubled. The maximum acceleration changes by a factor of

(A) $1/\sqrt{2}$

(B) $1/2$

(D) $1$

A mass $m$ on a vertical spring oscillates about its equilibrium position with period $T$ . At the instant the mass passes through equilibrium moving downward, a second identical mass is gently attached. Immediately after attachment, the new equilibrium position is

(A) unchanged

(B) lower by $mg/k$

(D) higher by $mg/k$

A pendulum bob of mass $m$ and length $L$ is also attached to a horizontal spring of constant $k$ that is relaxed when the bob hangs vertically. For small angles, compared with the same pendulum without the spring, the period is

(A) larger

(B) smaller

(D) zero because the forces cancel

A particle moves in the potential $U(x)=\dfrac{1}{2}kx^2+\epsilon x^4$ , where $k,\epsilon>0$ . For sufficiently small oscillations about $x=0$ , the angular frequency is

(A) $\sqrt{k/m}$

(B) $\sqrt{(k+4\epsilon)/m}$

(D) dependent on amplitude even in the small-amplitude limit

A particle moves near $x=0$ in the potential $U(x)=U_0+ax^2+bx^3+cx^4$ , where $a>0$ . For sufficiently small oscillations, the angular frequency is

(A) $\sqrt{a/m}$

(B) $\sqrt{2a/m}$

(D) $\sqrt{12c/m}$

A solid cylinder of mass $M$ and radius $R$ is attached at its center to a horizontal spring of constant $k$ and rolls without slipping. Its angular frequency is

(A) $\sqrt{k/M}$

(B) $\sqrt{2k/M}$

(D) $\sqrt{3k/(2M)}$

FRQ

A solid cylinder of mass $M$ and radius $R$ rests on a rough horizontal surface and rolls without slipping. A light spring of constant $k$ is attached to the cylinder’s center, and the other end is fixed to a wall. The cylinder is displaced a small distance $A$ from equilibrium and released from rest.

$(A)$ Using energy, derive an expression for the angular frequency of the oscillation in terms of $M$ and $k$ .

$(B)$ Determine the maximum static friction force needed during the motion.

$(C)$ Find the minimum coefficient of static friction required for rolling without slipping for the entire motion.

$(D)$ Suppose the cylinder is replaced by a thin hoop with the same $M$ and $R$ . Without redoing the full calculation, determine whether the period increases, decreases, or stays the same, and justify your answer.

A bead of mass $m$ slides without friction on a rigid circular wire of radius $R$ fixed in a vertical plane. The bead is also attached to a light spring of constant $k$ and negligible relaxed length whose other end is fixed at the top of the circle. Let $\theta$ be the bead’s angular displacement from the bottom of the circle.

$(A)$ Write the bead’s gravitational potential energy and spring potential energy as functions of $\theta$ , taking the bottom of the circle as zero gravitational potential.

$(B)$ Find the condition on $k$ and $R$ for the bottom of the circle to be a stable equilibrium.

$(C)$ For small oscillations about the bottom, derive the angular frequency in terms of $m$ , $g$ , $R$ , and $k$ .

$(D)$ Describe qualitatively how the equilibrium position changes if the spring constant is made very large.

A student studies a cart-spring oscillator on a horizontal track. The cart of mass $M$ has a small block of mass $m$ resting on top of it. The coefficient of static friction between the block and cart is $\mu_s$ . The cart is pulled to amplitude $A$ and released from rest; the block does not slip at first.

$(A)$ Derive the period of the combined motion while the block does not slip.

$(B)$ Determine the maximum amplitude $A_{\max}$ for which the block can remain at rest relative to the cart throughout the motion.

$(C)$ The student measures the period for several added top-block masses $m$ . Describe a graph that could be used to determine the spring constant $k$ from the data, including what should be plotted on each axis.

$(D)$ If the block begins to slip near the endpoints of the motion, explain whether the measured period should be expected to match the expression from part $(A)$ . Your explanation should refer to the forces on the two objects, not just energy loss.