Classical Mechanics - Harmonic Oscillator

A particle of mass m undergoes harmonic oscillation with period T₀. A force f proportional to the speed v of the particle, f = −bv, is introduced. If the particle continues to oscillate, the period with f acting is

A. Larger than T₀
B. Smaller than T₀
C. Independent of b
D. Dependent linearly on b
E. Constantly changing

(GR9677 #08)

Solution:

 f = −bv → minus sign means f is restoring force (damped oscillation).
The oscillation is getting slower (larger period) before it finally comes to stop.

Answer: A

Classical Mechanics - Moment Inertia

The period of physical pendulum is 2π√(I/mgd), where I is the moment of inertia about the pivot point and d is the distance from the pivot to the center of mass. A circular hoop hangs from a nail on a barn wall. The mass of the hoop is 3 kilograms and its radius is 20 centimeters. If it is displaced slightly by a passing breeze, what is the period of the resulting oscillations?

A. 0.63 s
B. 1.0 s
C. 1.3 s
D. 1.8 s
E. 2.1 s

(GR9677 #21)

Solution:

T = 2π√(I/mgd)
m = 3 kg
r = 20 cm = 0.2 m
In this case d = r

To find total I:
Parallel axis theorem: I = ml² + I_CM
I_CM  = I_loop = mr² 

In this case l = r
→ I = mr² + mr² = 2mr²

T = 2π√(2mr² / mgr) 
= 2π√(2r/g) 
= 2π√[2(0.2)/(10)] 
= 2π(0.2)
= 1.25 s

Answer: C

Classical Mechanics - Small Oscillation

Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diameter of X is also four times that of Y. If the period of small oscillations of X is T, the period of small oscillations of Y is

A. T
B. T/2
C. T/4
D. T/8
E. T/16

(GR9277 #74)

Solution:

Small oscillation → simple pendulum.

Period of simple pendulum:

→

In this case, = radius of the hoop.



Answer: B

Classical Mechanics - Wave equation

The equation where A, T, and λ are positive constants, represents a wave whose

A. Amplitude is 2A
B. Velocity is in the negative x–direction
C. Period is T/λ
D. Speed is x/t
E. Speed is λ/T

(GR8677 #04)

Solution:

(A) FALSE.
Amplitude  = maximum displacement = y_max = A.

(B) FALSE.
For a wave traveling to the right (positive x-direction): y = ƒ(x − vt)
(x − vt) = constant.
As t increases, x must increases to keep (x − vt) = constant

For a wave traveling to the left (negative x-direction): y = ƒ(x + vt)
As t increases, x decreases to keep (x + vt) = constant.

The problem gives:  y = ƒ(vt − x)
As t increases, x must increases to keep (x − vt) = constant.
The waves is traveling to the right.

(C) FALSE.
The unit of T/λ (second/meter) does not match with the unit is period (second).

(D) FALSE.
The speed of  the wave is λ/T.

(E) TRUE.
See D.

Answer: E

Classical Mechanics - Oscillatory Motion

Two identical springs with spring constant k are connected to identical masses of mass M, as shown in the figures. The ratio of the period for the springs connected in parallel (Fig. 1) to the period for the springs connected in series (Fig.2) is

A. ½
B. 1/√2
C. 1
D. √2
E. 2

(GR0177 #90)

Solution:

Spring constants:

Series, k_s = ½ k
Parallel, k_p = 2k

Period:




Answer: A

Classical Mechanics - Period of the Motion

A particle of mass m moves in the potential above. The period of the motion when the particle has energy E is

A.

B.

C.

D.

E.

(GR9677 #93)

Solution:

Total Period: T = T_SHO + T_grav

For V = ½kx² → Simple Harmonic Oscillator (SHO)
Period of SHO, T_SHO = 2π√(k/m)

The graph shows only half of the usual SHO potential:
T_SHO = ½ × 2π√(k/m) = π√(k/m)

Total Period: T = π√(k/m) + T_grav

Answer: D


Notes:

To find T_grav with V = mgx:

E = T + V = 0 + mgx 
x = E/mg

Kinematic Equation:

x = v₀t +  ½gt²
v₀ = 0
x = ½gt² = E/mg
t²= 2E/mg²
t = T_grav = √(2E/mg²)

Since the particle has to travel from the origin to the right endpoint and then back to the origin, the total time contribution from this potential:

T_grav = 2√(2E/mg²)

The total period is, T = π√(k/m) + 2√(2E/mg²)