PHY2604 Assignment 1 Memo 2024
1. Resonance
Resonance refers to the phenomenon where a system oscillates with the max-
imum possible amplitude at a given frequency. This frequency is known as
the resonance or natural frequency of the system and is determined by its
inherent properties.
2. Calculation
(a) Radius of the disc
diameter 0.135
R= = = 0.0675 m
2 2
(b) Moment of Inertia
For a disc rotating on its axis:
1
I = mR2
2
First, calculate the mass m of the disc:
m = ρV
V = area × thickness = πR2 × 0.02
V = π(0.0675)2 × 0.02 = 0.0002862763 m3
ρ = 8000 kg/m3
m = 8000 × 0.0002862763 = 2.29 kg
Now, calculate I:
1
I = (2.29)(0.0675)2 = 0.0052 kg · m2
2
1
,3. Force and Torque
(a) Force due to gravity
F = mg
F = 0.75 × 9.8 = 7.35 N
(b) Torque
τ = F d sin θ
τ = 7.35 × 0.33 × sin(90◦ )
τ = 2.4255 N · m
4. Frictional Torque
The friction of the rope on the pulley results in a torque force. Also, the
friction between the pulley and its axle results in a torque force opposing the
direction of movement.
5. Wavelengths and Damping
We notice 4 full wavelengths in 20 seconds.
Period T :
20
T = = 5s
4
Angular Velocity ω ′ :
2π
ω′ =
T
2π
ω′ = = 1.2566 rad · s−1
5
2
, Damping Constant γ:
From the equation (p. 35, Vibrations & Waves, George C. King):
At = A0 e−γt/2 ⇒ A0 = At eγt/2
Given:
A0 = 1, A2 = 0.5
A0 γT
ln =
A2 2
1 γ·5
ln =
0.5 2
γ·5
ln(2) =
2
2 ln(2)
γ= = 0.1386294361
5
Natural Angular Velocity ω0 :
r γ 2
ω= ω02 −
2
Rearranging for ω0 : γ 2
ω02 =ω + 2
2
Substitute:
γ 0.1386294361
ω = 1.2566, =
2 2
2
2 2 0.1386294361
ω0 = (1.2566) +
2
ω02 = 1.5834809
√
ω0 = 1.5834809 = 1.259 rad · s−1
3
1. Resonance
Resonance refers to the phenomenon where a system oscillates with the max-
imum possible amplitude at a given frequency. This frequency is known as
the resonance or natural frequency of the system and is determined by its
inherent properties.
2. Calculation
(a) Radius of the disc
diameter 0.135
R= = = 0.0675 m
2 2
(b) Moment of Inertia
For a disc rotating on its axis:
1
I = mR2
2
First, calculate the mass m of the disc:
m = ρV
V = area × thickness = πR2 × 0.02
V = π(0.0675)2 × 0.02 = 0.0002862763 m3
ρ = 8000 kg/m3
m = 8000 × 0.0002862763 = 2.29 kg
Now, calculate I:
1
I = (2.29)(0.0675)2 = 0.0052 kg · m2
2
1
,3. Force and Torque
(a) Force due to gravity
F = mg
F = 0.75 × 9.8 = 7.35 N
(b) Torque
τ = F d sin θ
τ = 7.35 × 0.33 × sin(90◦ )
τ = 2.4255 N · m
4. Frictional Torque
The friction of the rope on the pulley results in a torque force. Also, the
friction between the pulley and its axle results in a torque force opposing the
direction of movement.
5. Wavelengths and Damping
We notice 4 full wavelengths in 20 seconds.
Period T :
20
T = = 5s
4
Angular Velocity ω ′ :
2π
ω′ =
T
2π
ω′ = = 1.2566 rad · s−1
5
2
, Damping Constant γ:
From the equation (p. 35, Vibrations & Waves, George C. King):
At = A0 e−γt/2 ⇒ A0 = At eγt/2
Given:
A0 = 1, A2 = 0.5
A0 γT
ln =
A2 2
1 γ·5
ln =
0.5 2
γ·5
ln(2) =
2
2 ln(2)
γ= = 0.1386294361
5
Natural Angular Velocity ω0 :
r γ 2
ω= ω02 −
2
Rearranging for ω0 : γ 2
ω02 =ω + 2
2
Substitute:
γ 0.1386294361
ω = 1.2566, =
2 2
2
2 2 0.1386294361
ω0 = (1.2566) +
2
ω02 = 1.5834809
√
ω0 = 1.5834809 = 1.259 rad · s−1
3
