### Simple oscillations

#### Problem:

A particle that hangs from a spring oscillates with an angular frequency of 2 rad/s.  The spring is suspended from the ceiling of an elevator car and hangs motionless (relative to the car) as the car descends at a constant speed of 1.5 m/s.  The car then suddenly stops.  Neglect the mass of the spring.
(a) With what amplitude does the particle oscillate?
(b) What is the equation of motion for the particle?  (Choose the upward direction to be positive.)

Solution:

• Concepts:
Simple harmonic motion
• Reasoning:
After the elevator stops, we have a mass on a spring that moves through the equilibrium position with a given velocity.
• Details of the calculation:
(a) When traveling in the elevator at constant speed, the total force on the mass is zero.  The force exerted by the spring is equal in magnitude to the gravitational force on the mass, the spring has the equilibrium length of a vertical spring.  When the elevator suddenly stops, the end of the spring attached to the ceiling stops.  The mass, however has momentum, p = mv, and therefore starts stretching the spring.  It moves through the equilibrium position of the vertical spring with its maximum velocity vmax = 1.5 m/s.
Its velocity as a function of time is v(t) = -ωAsin(ωt + Φ).
Since vmax = ωA and ω = 2/s, the amplitude of the oscillations is A = 0.75 m.
(b) The equation of motion for the particle is d2x/dt2 = -(k/m)x = -ω2x.  Its solution is
x(t) = A cos(ωt + Φ) = (0.75 m) cos((2/s)t + Φ).
If we choose the t = 0 to be the time the elevator stops and let the upward direction be positive, then x(0) = 0, and v(0) = -1.5 m/s.  We therefore need Φ to be π/2.

#### Problem:

A mass of 250 g hangs on a spring and oscillates vertically with a period of 1.1 s.  What mass must be added to double the period?

Solution:

• Concepts:
Simple harmonic motion
• Reasoning:
A mass-spring system is a simple harmonic oscillator.
• Details of the calculation:
For simple harmonic motion T = 2π√(m/k).
T2/T1 = 2 = √(m2/m1).
m2/m1 = 4.
m2 = 1000 g.

#### Problem:

Two equal masses, each of mass m, are connected by a spring having a spring constant k.  If the equilibrium separation is L0 and the spring rests on a frictionless horizontal surface, what is the frequency of vibrations w in terms of k, m and L0?

Solution:

• Concepts:
Newton's second law
• Reasoning:
The force acting on each particle is given, F = -k(L – L0), where L is the length of the stretched or compressed spring.
• Details of the calculation:
Place the origin of the coordinate system at the center of mass.
For each mass we have:
F = -k(L – L0) = md2(½ (L-L0))/dt2.
d2(L-L0) /dt2= -(2k/m) (L – L0).
ω = (2k/m)½.

#### Problem:

A certain oscillator satisfies the equation d2x/dt2 + 4x = 0.  Initially the particle is at the point x = √3 when it is projected towards the origin with speed u = 2.
(a)  Find x(t).
(b)  How long does it take for the particle to first reach the origin?

Solution:

• Concepts:
Harmonic motion
• Reasoning:
Solve the differential equation for the given initial conditions.
• Details of the calculation:
(a)  Given:  d2x/dt2 + 4x = 0,  x(0) = 3,  (dx/dt)(0) = -2.
x(t) = A cos(2t + φ), dx/dt = -2A sin(2t + φ)
Initial conditions:  √3 = Acos(φ),  2 = 2A sin(φ)
tan(φ) = 1/√3,  φ = 30 deg = π/6,  A = √3/cos(π/6) = 2.
x(t) = 2 cos(2t + π/6)
(b)  x(t) = 0 --> cos(2t + π/6) = 0, 2t + π/6 = π/2, t = π/6 units = 0.524 units.

#### Problem:

A simple pendulum is attached to the ceiling of a boxcar which accelerates at a constant rate a.  Find the equilibrium angle of the pendulum, and also the frequency of small oscillations.

Solution:

• Concepts:
Motion in an accelerating frame, small oscillations
• Reasoning:
The boxcar is not an inertial frame.
• Details of the calculation:

The pendulum bob is in equilibrium in the accelerating frame.  It is accelerating in an inertial frame with acceleration a = ai.  The forces on the bob are Fg = mg and T due to the tension in the string.  We have:
Tcosθ = mg,  Tsinθ = ma, tanθ = a/g.
An observer in the boxcar experiences a fictitious force -mai.  The total force observed in the accelerating frame is therefore -m(gj + ai).  The magnitude of this force is mg', with g' = g/cosθ. or g' = (g2 + a2)½.  Small angular displacements δ from the equilibrium angle θ will result in a restoring force proportional to δ.  d2δ/dt2 = -(g'/l)δ.  The frequency of small oscillations is ω = (g'/l)½.

#### Problem:

Calculate the frequencies of oscillations of the mass m for the two spring configurations shown in the figures.  The springs have elastic constants k1 and k2.

Solution:

• Concepts:
Simple harmonic motion
• Reasoning:
We have motion in 1 dimension of a a single mass connected to springs.
• Details of the calculation:
(a)  F = - k1x – k2x = md2x/dt2,  ω2 = (k1 + k2)/m,  f = (k1 + k2))½/(2π m½).
(b)  F = - k2 x2,  k1x1 = k2 x2,  x1 + x2 = x.  x2 = k1x/(k1 + k2).
F = - k2k1x/(k1 + k2) = md2x/dt2,  ω2 = k2k1/[m(k1 + k2)],  f = [k2k1/(k1 + k2)]½/ ((2π m½).

#### Problem:

When the system shown in the diagram is in equilibrium, the right spring is stretched by x1.  The coefficient of static friction between the blocks is µs.  There is no friction between the bottom block and the supporting surface.  The force constants of the springs are k and 3k (see the diagram).  The blocks have equal mass m.  Find the maximum amplitude of the oscillations of the system shown in the diagram that does not allow the top block to slide on the bottom.

Solution:

• Concepts:
Simple harmonic motion
• Reasoning:
As long as the blocks stick together, they will execute simple harmonic motion about an equilibrium position.
For a displacement x from the equilibrium position, the force on each block then will obey Hooke's law, F = -αx.
We can find the equilibrium position and α, since we know the spring constants.
But the force on each block is also the sum of the force exerted on the block by the spring it is attached to and the frictional force.  Equating the two expressions for the force we can find an expression for the frictional force f, and setting if equal to its maximum value we can find the maximum allowed amplitude for the blocks to not start sliding with respect to each other.
• Details of the calculation:
Let displacements to the right be positive and displacements to the left be negative.
In equilibrium:
Force on upper block = -3kx2 + f = 0.  Force on lower block = -kx1 - f = 0.
x2 = -x1/3  (Note: x1 is negative, x2 is positive.)  The left spring is stretched by x1/3.
Assume the masses are stuck together.  What is the equation of motion?
Let x be their displacement from equilibrium towards the right.
-kx - 3kx = 2md2x/dt2.  md2x/dt2 = -2kx.  x = Acosωt.  ω2 = 2k/m.
The net force on each block therefore is given by F = ma = -2kx.
But we also have for the upper block F = -3k(x + x2) + f.
Therefore f = kx + 3kx2
|fmax| = kA + 3kx2 = kA + k|x1| during one cycle of the motion, when x = A.
For the lower block we have F = -k(x + x1) + f, therefore f = -kx + kx1.
|fmax| = kA + k|x1| during one cycle of the motion, when x = -A.
If we set |fmax| = µsmg, we have Amax = µsmg/k - |x1|.

#### Problem:

Three small identical coins of mass m each are connected by two light non-conducting strings of length d each.  Each coin carries an unknown charge q.  The coins are placed on a horizontal frictionless non-conducting surface as shown (the angle between the strings is very close to 180°).  After the coins are released, they are observed to vibrate with period T.  Find the charge q on each of the coins in terms of m, d, and T.

Solution:

• Concepts:
Small oscillations
• Reasoning:
The CM has to remain stationary.  Choose a coordinate system as shown in the figure below.

The system can only oscillate as indicated by the arrows in the figure.
• Details of the calculation:
Let us find the force on mass 3 located at x = 0 as a function of its distance from the CM.
Fx = 0, Fy = 2Ftsinθ - 2kq2sinθ/d2.
To find the tension Ft we calculate the component of the force, Fr, on mass 2 along the direction making an angle θ with the x-axis.  We need Fr = 0, since the string has a fixed length.
Fr = kq2/d2 + kq2cosθ/(2dcosθ)2 – Ft = 0.  Ft = kq2/d2 + kq2cosθ/(2dcosθ)2.
In the small angle approximation sinθ = θ and cosθ = 1.
Ft = (5/4)kq2/d2.
Therefore Fy = ½kq2θ/d2.  Fy is proportional to θ.
The relationship between y and θ is y = -⅔dsinθ = -⅔dθ, in the small angle approximation.  (The CM lies ⅔dsinθ above mass 3.)
Therefore Fy = -¾kq2y/d3.  Fy = -ay.  Fy obeys Hooke's law.
The mass oscillates with angular frequency ω = (a/m)½ = (3kq2/(4md3))½ .
The period of oscillation is T = 2π/ω.
T2 = 16π2md3/(3kq2) = 64π3ε0md3/(3q2). q2 = 64π3ε0md3/(3T2).
q = (64π3ε0md3/(3T2))½.

#### Problem:

Consider the motion of a point of mass m subjected to a potential energy function of the form
U(x) = U0[1 - cos(x/R0)] for πR0/2 < x < πR0/2,
where x denotes distance, and U0 and R0 are positive constants with dimensions of energy and length, respectively.
(a)  Find the position of stable equilibrium for the mass.
(b)  Show that the motion of the mass in proximity of the stable equilibrium position is SHM.
(c)  Find the period of the small oscillations.
(d)  Find the period of the small oscillations for the same mass in the potential
U(x) = -U0/[1 + (x/R0)2].

Solution:

• Concepts:
• Reasoning:
We are asked to find the period of small oscillations.
• Details of the calculation:
(a)  The position of stable equilibrium is x = 0.
dU/dx = (U0/R0)sin(x/R0),  d2U/dx2 = (U0/R02)cos(x/R0).
dU/dx|0 = 0.  d2U/dx2|0 > 0.
(b)  For x << R0 we have U(x) = ½(d2U/dx2)|0x2.  This is the potential energy of a harmonic oscillator U(x) = ½kx2, with k = d2U/dx2|0
(c)  ω2 = (1/m)(d2U/dx2)|0 = U0/(mR02).  T = 2π/ω = 2πm1/2R0/U01/2.
(d)  dU/dx = (U0/R02)2x/(1 + x2/R02)2,  dU/dx|0 = 0, x = 0 is an equilibrium position.
d2U/dx2|0 = (2U0/R02) > 0.
For x << R0, U(x) = -U0(1 – x2/R02) = -U0  + (U0/R02)x2.
ω2 = (2U0/(mR02)).  T = 2π/ω = 21/2πm1/2R0/U01/2.

#### Problem:

A flexible “U-tube” with cross sectional area A is filled with a liquid of density ρ that has no viscosity.  The total length of the fluid column is L.  If one side of the U tube is suddenly lowered so that the fluid level on that side is lowered by an amount h with respect to its initial position, when will the heights of the two columns again be equal?  (You may assume that the fluid flow is laminar but with no sticking to the inner surface of the U tube so that all particles in the fluid move at the same speed.  Also, neglect all friction, such as any effects of air resistance in the tube above the liquid.)

Solution:

• Concepts
The harmonic oscillator
• Reasoning:
The fluid in the tube will oscillate about its equilibrium position.
• Details of the calculation:

Height difference when the fluid on one side is lowered by an amount y:  2y
Magnitude of net force pushing the fluid towards its equilibrium position:  Fy = -2ρgAy
ρALd2y/dt2 = -2ρgAy,  d2y/dt2 = -(2g/L)y,  y = h cos(ωt),  ω = (2g/L)½, given the initial conditions.
When ωt = π/2,   t = (π/2)*(L/(2g))½, the heights of the two columns will again be equal.