A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is

Total distance covered by the particle\\( = 4A\\). we divide this whole path in 8 intervals of \\(A/2\\). so, 5/8 oscillations means, it has already completed\\(1/2\\) oscillation(i.e. total dist. = 2A) and is half way to the other side i.e. A/2. so,\\( A/2 = Asinwt, w= 2pi/ T\\) , substitute to get \\(t= T/12.\\) now, total time taken = time to complete previous one half(2A) + time taken to complete\\( A/2 = t/2 + t/12 = 7T/12.\\)

The potential energy of a particle of mass \\(1\ kg\\) in motion along the \\(x\\)-axis is given by \\(U = 4(1 - \cos 2x)\ J\\). Here \\(x\\) is in meter. The period of small oscillations (in sec) is _______.

\\(M=1\ kg\\)\\(U=4(1-\cos2 x)\\)\\(F=-\dfrac {du}{dx}\\)\\(F=-4 [0-(-\sin 2x) (2)]\\)\\(F=-8\sin 2x\ N\\)\\(a=-8\sin 2x \ m/ \sec^2 \quad [\because \ m=1\ kg]\\)\\(\Rightarrow \ a\simeq -8(2x)\\)\\(a=-16x\\) [For small \\(x, \sin x \approx x\\)]\\(\Rightarrow \ \omega^2 =4^2\\)\\(\omega=4\ \\) rad /ac\\(T=\dfrac {2\pi}{w}=\dfrac {\pi}{2}\sec\\)

The damping force on a oscillator is directly proportional to the velocity. The Unit of the constant of proportionality are:

\\(F=Kv\\)\\(K=\dfrac{F}{V}\\)Unit of K = \\(K=\dfrac{kg ms^{-2}}{ms^{-1}}=kgs^{-1}\\)

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be \\(\displaystyle\dfrac{1}{X}\\) times the original, where \\(X\\) is

Amplitude of damped oscillator\\(A=A_0e^{- \lambda t} ; \lambda =constant, t=time\\)For \\(t=1\ min. \dfrac{A_0}{2}=A_0 e^{- \lambda t} \Rightarrow e^{\lambda}=2\\)For \\(t=3\ min.\ A=A_0e^{- \lambda \times 3 }=\dfrac{A_0}{(e^{\lambda})^3}=\dfrac{A_0}{2^3}\\)\\(\Rightarrow X=2^3\\)

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to \\(\dfrac{1}{1000} \\) of the original amplitude is close to :-

\\(A = A_0 e^{-\gamma t}\\)\\(A = \dfrac{A_0}{2}\\) after 10 oscillations\\(\because \\) After 2 seconds\\(\dfrac{A_0}{2} = A_0 e^{-\gamma (2)}\\)\\(2 = e^{2 \gamma}\\)\\(\ell n 2 = 2 \gamma \\)\\(\gamma = \dfrac{\ell n 2}{2}\\)\\(\because A = A_0 e^{-\gamma t}\\)\\(\ell n \dfrac{A_0}{A} = \gamma t\\)\\(\ell n 1000 = \dfrac{\ell n 2}{2} t\\)\\(2 \left(\dfrac{3 \ell n 10}{\ell n2} \right) = t\\)\\(\dfrac{6 \ell n 10}{\ell n 2} = 1\\)\\(t = 19.931 sec\\)\\(t \approx 20 \, sec\\)

A particle of mass \\('m'\\) is attached to three identical springs \\(A, B\\) and \\(C\\) each of force constant \\('k'\\) as shown in figure. If the particle of mass \\('m'\\) is pushed sightly against the spring \\('A'\\) and released. Find the period of oscillation.

\\({F_{net}} = {F_A} + {F_B}\cos {45^0} + {F_C}\cos {45^0}\\)\\( = ky + 2ky'\cos {45^0} = ky + 2k\left( {y\cos {{45}^0}} \right)\cos {45^0} = 2ky\\)\\({F_{net}} = k'y \Rightarrow k'y = 2ky \Rightarrow k' = 2k\\)\\(T = 2\pi \sqrt {\dfrac{m}{{k'}}} = 2\pi \sqrt {\dfrac{m}{{2k}}} \\)we get, period of oscillation is \\(2\pi \sqrt {\dfrac{m}{{2k}}} .\\)

Resonance is a special case of forced vibration in which the natural frequency of vibration of the body is the same as the impressed frequency of external periodic force and the amplitude of forced vibration is maximum.

The amplitude of forced vibrations of a body increases with an increase in the frequency of the externally impressed periodic force.

A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity \\(\sqrt{\dfrac{7gL}{2}}\\). The thread will slack after swinging through an angle \\(\theta\\), such that \\(\theta\\) equal.

As shown in figure, \\(h=L+\sin\phi\\)By energy conservation,\\(\dfrac{1}{2}mv^2=mgh+\dfrac{1}{2}mv'^2\\)\\(\dfrac{1}{2}\times (\sqrt{\dfrac{7gl}{2}})^2=g(L+L\sin\phi)+\dfrac{1}{2}v'^2\\)\\(v'^2=-2gL-2gL\sin\phi+\dfrac{7}{2} gL\\)\\(=\dfrac{3}{2}gL-2gL\sin\phi\\)If string slack then,\\(mg\sin\phi=\dfrac{mv'^2}{L}\\)\\(g\sin\phi=\dfrac{1}{L}(v')^2=\dfrac{1}{L}(\dfrac{3}{2}gL-2gL\sin\phi)\\)\\(g\sin\phi=\dfrac{3}{2}g-2g\sin\phi\\)\\(3g\sin\phi=\dfrac{3}{2}g\\)\\(\sin\phi=\dfrac{1}{2}\\)\\(\phi=\dfrac{\pi}{6}\\)\\(\theta=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{4\pi}{6}=120^\circ\\)

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be :

For a spring-mass damped oscillator\\(\dfrac{d^2x}{dt^2}+\left(\dfrac{2x}{2\sqrt{mk}}\sqrt{\dfrac{k}{m}}\right)\dfrac{dx}{dt}+\left(\sqrt{\dfrac{k}{m}}\right)^2x=0\\)\\(\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+2\tau w_0\dfrac{dx}{dt}+w_0^2x=0}\\)\\(\tau=\dfrac{c}{2\sqrt{mk}}\Rightarrow \\) damping ratio \\(c=b\Rightarrow\\) damping constant\\(W_0=\sqrt{\dfrac{k}{m}}\Rightarrow\\) angular frequency\\(\Rightarrow \boxed{\dfrac{d^2x}{dt^2}+(\dfrac{b}{m})\dfrac{dx}{dt}+(\dfrac{k}{m})x=0}\\)..........(1)For LCR circuituse KVL:-\\(V-L\dfrac{dI}{dt}-IR-\dfrac{q}{c}=0\\)\\(\Rightarrow \boxed{\dfrac{d^2q}{dt^2}+\dfrac{R}{L}\dfrac{dq}{dt}+(\dfrac{1}{LC})q=\dfrac{V}{L}}\\)...........(2) \\(I=\dfrac{dq}{dt}\\)By comparing (1) and (2):=\\(L\rightarrow m,\ \ \ R\rightarrow b,\ \ \ \ \ LC\rightarrow \dfrac{m}{k}\\) \\(\Rightarrow C\rightarrow \dfrac{1}{k}\\)option (D) is correct.

In forced oscillation of a particle the amplitude is maximum for a frequency \\(\omega_1\\) of force, while the energy is maximum for a frequency \\(\omega_2\\) of the force, then:

For the amplitude of oscillation and energy to be maximum, the frequency of force must be equal to the initial frequency and this is only possible in resonance. In resonance state \\( \omega_1 = \omega_2\\).

Free and Forced Oscillations | Formulas, Definition, Examples

formula

Find time period of oscillation assuming small damping

ω = (\frac{k}{m} - \frac{r ^{2}}{4 m ^{2}})

w = \frac{2 π}{T}

example

Find external force in forced oscillation

Example: A simple harmonic oscillator is of mass

0.100 k g

. It is oscillating with a frequency of

\frac{5}{π}

Hz. If its amplitude of vibration is

5

cm, what is the force acting on the particle at its extreme position?

Solution:

F = m w^{2} A

ω = 2 π f = 10 r a d / s e c

F = 0.1 \times 100 \times \frac{5}{1 0 0}

= 0.5 N

example

Frequency of oscillations in forced oscillations

Example: A

2

kg block hangs without vibrating at the bottom end of a spring with a force constant of

400

N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of

5

m/s

^{2}

when the acceleration suddenly ceases at time

t = 0

and the car moves upward with constant speed (

g = 10

m/s

^{2}

) What is the angular frequency of oscillation of the block after the acceleration ceases?

Solution:
Given : Mass of the block

m = 2 k g

Spring constant

k = 400 N / m

Let angular frequency of the oscillation be

ω

.
Now using

ω = \frac{k}{m}

ω = \frac{4 0 0}{2} ⟹ ω = 102 r a d / s

definition

Write displacement as a function of time in forced oscillation

The object oscillates about the equilibrium position

x_{0}

. If we choose the origin of our coordinate system such that

x_{0}

= 0

, then the displacement

x

from the equilibrium position as a function of time is given by:

x (t) = A c o s (w t + ϕ)

example

Examples of mechanical resonance

Various examples of mechanical resonance include:
     1. Musical instruments (acoustic resonance).
   2. Most clocks keep time by mechanical resonance in a balance wheel,   pendulum, or quartz crystal.
   3. Tidal resonance of the Bay of Fundy.
   4. Orbital resonance as in some moons of the solar system's gas giants.
   5. The resonance of the basilar membrane in the ear.
   6. Making a child's swing, swing higher by pushing it at each swing.
   7. A wine glass breaking when someone sings a loud note at exactly the right   pitch.

definition

Oscillations when driving frequency is close to natural frequency

Amplitude of oscillations in shown in the attached plot and given by the formula:

A = \frac{F _{o}}{m _{0}^{2} ( ω ^{2} - ω _{d}^{2} ) ^{2} + ω _{d}^{2} b ^{2}}

where

F_{o} :

Driving Force

m_{o} :

Mass

ω :

Driving Frequency

ω_{d} :

Damping Frequency and

b :

Damping constant

When

ω = ω_{d}

, amplitude of oscillations is maximum.

A = \frac{F _{o}}{ω _{d} b}

which is a very large value.

This is the phenomenon of resonance in forced oscillations.

diagram

Displacement time graph for different oscillations

definition

Explain maintained oscillation with example

Maintained Oscillation : In maintained oscillation, energy is supplied to the system from outside at the same rate at which the energy is lost by it, so that its amplitude remains constant & the system oscillates with its own natural frequency. Example: Swing, electric tuning fork, balance wheel of a watch, etc

Free and Forced Oscillations

formula

Find time period of oscillation assuming small damping

example

Find external force in forced oscillation

example

Frequency of oscillations in forced oscillations

definition

Write displacement as a function of time in forced oscillation

example

Examples of mechanical resonance

definition

Oscillations when driving frequency is close to natural frequency

diagram

Displacement time graph for different oscillations

definition

Explain maintained oscillation with example

REVISE WITH CONCEPTS

Damped SHM

Differential Equation of Motion in Forced Oscillations

Effect of Damping on Resonance

Quick Summary With Stories