a) A simple harmonic motion is represented by x(t) = a cosωt Obtain expressions for velocity and acceleration of the oscillator. Also, plot the time variation of displacement, velocity and acceleration of the oscillator. b) The time period of a simple pendulum, called ‘seconds pendulum’, is 2 s. Calculate the length, angular frequency and frequency of the pendulum. What is the difference between a simple pendulum and a compound pendulum? (6+4) c) Two collinear harmonic oscillations x1 = 8 sin (100 πt) and x2 = 12 sin (96 πt) are superposed. Calculate the values of time when the amplitude of the resultant oscillation will be (i) maximum and (ii) minimum. (5+5) d) For a damped harmonic oscillator, the equation of motion is ( / ) ( / ) 0 2 2 m d x dt + γ dx dt + kx = with m = 0.50 kg, γ = 0.70 kgs−1 and k = 70 Nm−1. Calculate (i) the period of motion, (ii) number of oscillations in which its amplitude will become half of its initial value, (iii) the number of oscillations in which its mechanical energy will drop to half of its initial value, (iv) its relaxation time, and (v) quality factor. (4+4+4+4+4) e) Establish the equation of motion of a weakly damped forced oscillator explaining the significance of each term. Differentiate between transient and steady state of the oscillator.
Answers
a) x(t) = a
cos ωt
v = dx/dt = - a ω sinωt
a = dv/dt = - aω² Cosωt = - ω²
x
b) T = sec Pendulum clock
T = 2π √(L/g)
=> L = T² g /(4π²)
L = 2² * 10 /(4π²) = 1.013
meter
f = 1/T = 0.5 Hz
ω = 2πf = π rad/s
Simple pendulum has a point mass m
suspended by a light thread and the other end is tied fixed. The length of the
thread is L.
Compound pendulum has a mass
throughout the object that is suspended by hanging at a point on object.
A rod having mass fixed to a wall at the top end (optionally carrying
pendulum at the other end) is a compound pendulum. So in real life
all penduli are compound penduli.
c) Collinear Harmonic oscillators... in SHM are
superposed.
x1 = 8 sin (100π t)
x2 = 12 sin (96π t)
x = x1+x2 = 8 sin (100π t) + 8
Sin (96π t) + 4 Sin (98πt - 2πt)
= 8 *2
Sin(98πt) Cos(2πt) + 4 Sin 98πt Cos 2πt - 4 Cos 98πt sin 2πt
= 20 Sin (98πt)
cos (2πt) - 4 cos(98πt) Sin2πt
Here we can see that the result is
superposition of two enveloped waves. The
envelope of the first wave is cos 2πt
and Sin2πt for the second wave. Because
the difference in frequency 98π
and 2π is very large, ignore the larger frequency.
Effectively we have,
x(t) = 20 Cos (2πt) - 4
Sin 2πt
=> x'(t) = - 40π Sin
2πt - 8π Cos 2πt = 0
=>
Tan 2πt = - 1/5 => 2πt = (π -1/5) + nπ
for integer n
t = 1/2π (π -
1/5) + n/2
There are the
instants where the superposed wave has maxima or minima.
Max
x(t) = 20 cos (2π - 1/5) - 4 Sin (2π - 1/5) = 20.394
when 2 πt = 2 π - 0.2
rad
Minimum = 0 when Tan 2πt = 5.
=================
d)
A damped harmonic oscillator: Equation of motion :
x = displacement, t = time, k = constant (like
spring constant)
¥ = damping factor (drag coefficient) = 0.70 N/m,
m = mass of particle executing Periodic/oscillations = 0.50 kg
The value of k is missing... Let us assume k = 70 N/m
m d²x/dt² + ¥ dx/dt + k x = 0
dx²/dt² + 1.40 dx/dt + 140 x = 0
std form of ODE: d²x/dt² + p(t)
dx/dt + q(t) x = 0
Using the method of solving Ordinary differential equations for
second degree equations:
Let x(t) = u(t) v(t)
p(t) = 1.40, v(t) = exp(-0.70 t), q(t) = 140
Q(t) = v''(t) + p(t) v'(t) + q(t) v(t) = 140 exp(- 0.70 t)
Normal form of ODE: v(t) u''(t) + Q(t) u(t) = 0
u''(t) +140 u(t)
= 0
This is the equation of motion for a SHM.
ω² = 140, ω = 2√35 rad/s
u(t) = A Sin ωt
Solution : x(t) = A exp(- 0.70 t) Cos ωt
If we are given x(t=0), then we can know the value of A.
Period = 2π/ω =π/√35 Sec = 0.531 sec
Amplitude = A exp(- 0.70 t)
Time duration for amplitude to be come 1/2 of initial value
=> exp(- 0.70 t) = 1/2
=> t = Ln 2 / 0.70 = 0.99 sec
Number of oscillations = 0.99 /0.531 = 1.86
mechanical energy of the oscillator
= 1/2 k A² = 1/2 * 70 * A² * exp(- 1.40 t)
It becomes half in t = Ln 2 /1.40 = 0.495 sec
ie., in 0.93 oscillations.
Relaxation time period of a damped oscillator is the time duration for its
amplitude become 1/e of its initial value:
So relaxation time => 0.70 t = 1,
t = 1.43 sec
Quality factor = Energy stored in the oscillator / Energy lost during one
oscillation
= energy stored in the oscillator
/energy lost during one radian of oscillation
Q = ω/p(t) = 2√35 / 1.40 = 8.45
Here p(t) = gamma is less than 2 ω. So it
is underdamping.