Chapter 2 - Damped Simple Harmonic Motion

Question 1

What would Newton’s Second law om motion become if a frictional force acts in the direction opposite to the velocity?

Answer 1

mx^.. = - sx - rx^.

or

mx^.. + rx^. + sx = 0

Question 2

How would you derive the equation of motion from Newton’s second law when a frictional force is present?

Answer 2

1. mx^.. + rx^. + sx = 0
2. No 1, will have a solution of the form x = Ce^αt
3. Then x^. = αCe^αt and x^.. = α²Ce^αt
4. From 1 and 3 we have: Ce^αt(mα² + rα + s) = 0
5. So either Ce^αt or (mα² + rα + s) is equal to 0
6. If (mα² + rα + s) = 0 then:

α = (-r/2m) ± (r²/4m² – s/m)^1/2

7. Substituting 6 into x = Ce<sup>αt </sup>gives the equation.

Remember that x = x₁ + x₂

Question 3

We know that the equation of displacement is given by:

x = Ce^αt

where

α = (-r/2m) ± (r²/4m² – s/m)^1/2

Discuss the different types of damping.

Answer 3

The bracket (r²/4m² – s/m) could be positive, zero or negative, resulting in different types of damping:

1. (r²/4m² – s/m) > 0, The damping resistance term r²/4m² dominates the stifffness term s/m, and heavy damping results in a dead beat system.
2. (r²/4m² – s/m) = 0, The balance between the two terms in results in a critically damped system. (Reaches x = 0 in minimum time).
3. (r²/4m² – s/m) < 0, The system is lightly damped and gives oscillatory damped simple harmonic motion. (Sometimes called underdamped).

Question 4

Give the equation for damped simple harmonic motion.

Answer 4

x = Ae^-rt/2msin(ω’t + ϕ)

Where:

• A and ϕ are constants which depends on the motion at t = 0. The constant A is the value to which the amplitude would have risen at the first maximum if no damping were present.
• The displacement varies sinusoidaly with time as in the case of SHM, but has a new frequency:

ω’ = (s/m – r²/4m²)^1/2

Question 5

Give the equation for the logarithmic decrament.

Answer 5

Start with:

A₀/A₁ = e^rτ’/2m = e^δ

so that

δ = r/2m τ’ = Log_e A₀/A₁

Question 6

Define the relaxation time or modulus of decay.

Answer 6

1. It is the time taken for the amplitude to decay to:

e^-1 = 0.368

of its original value.

2. The amplitude is:

A_t = A₀e^rt/2m = A_oe^-1

at a time t = 2m/r

Question 7

Give the different ways of describing the damping of an oscillator.

Answer 7

1. Logarithmic decrament:

δ = r/2m τ’ = Log_e A₀/A₁

1. Relaxation time or modulus of decay:

A_t = A₀e^rt/2m = A_oe^-1

1. The quality factor or Q-value:

Energy stored in system = Q

Energy lost per cycle 2π

1. Energy dissipation :

dE/dt = -rx^.2

Question 8

Define the quality factor or q-Value.

Answer 8

1. This measures the rate at which the energy decays.
2. The quality factor is defined as:

Q = ω’m/r

as the number of radians through which the damped system oscillates as its energy decays to:

E = E₀e^-1

1. Q could be described as the number of oscillations before the oscillations become imperceptible (0).

Question 9

Give the voltage equation when there is resistance.

Answer 9

L(dI/dt) + RI +q/c = 0

Question 10

Give the equation for q.

Answer 10

q = q_oe^{-Rt/2L ± (R2/4l2 – 1/LC)1/2t}

Remeber to the power two’s not 2…..