Classical Mechanics - Pendulum

Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths l, but the pendulum balls have unequal masses m₁ and m₂. The initial distance between the masses is the equilibrium length of the spring, which has spring contant K. What is the highest normal mode frequency of this system?

A.

B.

C.

D.

E.

(GR9677 #84)

Solution:

A. FALSE
ω = √(g/l) is angular frequency for single pendulum

B. and C are FALSE
Both answers do not depend on g/l

D. TRUE
If there is no dependence on K → ω = √(g/l)
And if m₂ → ∞,  m₁ will still oscillate with spring constant K but has no dependence on m_2, as if it were connected to a stationary object.

E. FALSE
If there is no dependence on K → ω = √(2g/l), not angular frequency for single pendulum
And if m₂ → ∞,  ω = √(2g/l) has no dependence on K and m₁

Answer: D 

Classical Mechanics - Simple Harmonic Oscillation

Three masses are connected by two springs as above. A longitudinal normal mode with frequency ¹/_2π√^k/_m is exhibited by

1. A, B, C all moving in the same direction with equal amplitude.
2. A and C moving in opposite, and B at rest.
3. A and C moving in the same direction with equal amplitude, and B moving in the opposite direction with the same amplitude.
4. A and C moving in the same direction with equal amplitude, and B moving in the opposite direction with twice the amplitude.
5. None of the above.

(GR8677 #43)

Solution:

Angular frequency, ω = 2πf  → f = ^ω/_2π
For Simple Harmonic Oscillation (SHO), ω = √^k/_m
Thus, f = ¹/_2π√^k/_m is frequency of SHO.

SHO is a periodic motion of a particle about a fixed point (the centre of motion).

Only choice (B) gives a fixed point (mass B at rest).

Answer: B

Classical Mechanics - Harmonic Oscillator

Two identical blocks are connected by a spring. The combination is suspended, at rest, from a string attached to the ceiling, as shown in the figure. The string breaks suddenly. Immediately after the string breaks, what is the downward acceleration of the upper block?

A. 0
B. g/2
C. g
D. √2 g
E. 2g

(GR0177 #72)

Solution:

In equilibrium (before the string is cut):
Block 1:

                                    …(Eq.1)

Block 2:
                  … (Eq.2)

After the string is cut:
, and
(minus sign    is because the direction is in downward or parallel with  )

(Eq.1) becomes:

                                                                   ...(Eq.3)

Submit (Eq.2) to (Eq.3):   

Since  

Answer: E

Classical Mechanics - Hamiltonian

Two small equal masses m are connected by an ideal massless spring that has equilibrium length l₀ and force constant k. The system is free to move without friction in the plane of the page. If p₁ and p₂ represent the magnitude of momenta of the two masses, a Hamiltonian for this system is.

A. 

B.  

C.  

D.  

E.  

(GR0177 #92)

Solution:

Hamiltonian: H = T + U

T_i = p_i²/2m
T₁ + T₂ =  p₁²/2m + p₂²/2m

U = ¹/₂k(Δl)² = ¹/₂k(l − l₀)²

H = p₁²/2m + p₂²/2m + ¹/₂k(l − l₀)²
= ¹/₂[ p₁²/m + p₂²/m + k(l − l₀)²]

Answer: E