(i) When amplitudes of the vibrating pendulum is small then pendulum is small. Here the restoring force F = mg sin θ = mg θ = mgx/l .
(ii) Where x is the displacement of the bob and I is the length of pendulum. Hence F x x. Since F is directed towards mean position.
(ill) Therefore the motion of the bob of simple pendulum will be S.H.M. if θ is small.

When a piece of wood is pressed and released,
\(\mathrm{F}=\mathrm{ma}, \quad \mathrm{m} =\text { volume } \times \text { density }=\mathrm{A} \times \rho  \)
\(\text { Change in force } =\mathrm{mg}=\mathrm{A} \times \rho \mathrm{g}  \)
\(\therefore \text { Acceleration a } =\frac{F}{m}  \)
\(a =\left(\frac{A \rho g}{m}\right) x  \)   .....(1)
For SHM, \(a =\omega^{2} x\)   .....(2)
From equation (1) & (2) we get
\(\omega^{2}=\frac{A \rho g}{m} \quad \therefore \omega=\sqrt{\frac{A \rho g}{m}}\)
Time period \(\mathrm{T}=\sqrt{\frac{2 \pi}{\omega}} \quad \therefore \mathrm{T}=2 \pi \sqrt{\frac{m}{A \rho g}}\)

Frequency of simple harmonic motion is defined as the number of oscillations produced by the Particle.

Kinetic energy is KE\(=\frac { 1 }{ 2 } { mv }_{ x }^{ 2 }\)
Multiply numerator and denominator by m
\(KE=\frac { 1 }{ 2m } { m^{ 2 }v }_{ x }^{ 2 }=\frac { 1 }{ 2m } \left( { mv }_{ x } \right) ^{ 2 }=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }\)
where, P_x is the linear momentum of the particle executing simple harmonic motion.
Total energy can be written as sum of kinetic energy and potential energy, therefore, from equation (10.73) and also from equation (10.75), we get
E = KE + U(x) \(=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }+\frac { 1 }{ 2m } { m\omega ^{ 2 }v }_{ x }^{ 2 }\) = constant

When the mass collides with the spring, from the law of conservation of energy “the loss in kinetic energy of mass is gain in elastic potential energy by spring”.
Let x be the distance of compression of spring, then the law of conservation of energy
\(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } kx^{ 2 }\Rightarrow x=v\sqrt { \frac { m }{ k } } \)

Simple harmonic oscillation equation is
y = A sin(\(\omega\)t + \(\varphi_{0}\)) or y = A cos(\(\omega\)t + \(\varphi_{0}\))
a. For the wave, y = 0.3 sin(40\(\pi\)t +1.1)
Amplitude is A = 0.3 unit
Angular frequency \(\omega\) = 40\(\pi\) rad s⁻¹
Frequency f = \(\frac { \omega }{ 2\pi } =\frac { 40\pi }{ 2\pi } =20Hz\)
Time period T= \(\frac { 1 }{ f } =\frac { 1 }{ 20 } =0.05s\)
Initial phase is \(\varphi_{0}\) = 1.1 rad
b. For the wave, y = 2cos(\(\pi\)t)
Amplitude is A = 2 unit
Angular frequency \(\omega\) = \(\pi\) rad s⁻¹
Frequency f = \(\frac { \omega }{ 2\pi } =\frac { \pi }{ 2\pi } =0.5Hz\)
Time period T= \(\frac { 1 }{ f } =\frac { 1 }{ 0.5 } =2s\)
Initial phase is \(\varphi_{0}\) = 0 rad
c. For the wave, y = 3 sin(2\(\pi\)t + 1.5)
Amplitude is A = 3 unit
Angular frequency \(\omega\) = 2\(\pi\) rad s⁻¹
Frequency f =\(\frac { \omega }{ 2\pi } =\frac { 2\pi }{ 2\pi } =1Hz\)
Time period T= \(\frac { 1 }{ f } =\frac { 1 }{ 1 } =1s\)
Initial phase is \(\varphi_{0}\) = 1.5 rad

Amplitude at any instant is given by
\(a=a_{0} e^{-b t}\)
\(\mathrm{a}_{0}=\text { initial amplitude }\)
b = damping constant
Case 1:
\(t =100 \mathrm{~T} \)
\(a =\frac{a_{0}}{3} \)
\(\therefore \frac{a_{0}}{3} =a_{0} e^{-b 100 T} \)
\(e^{-b 100 T}=\frac{1}{3}\) ....(1)
Case 2:
\(a=a_{0} e^{-200 b T} \)
\(a=a_{0}^{(-100 b T) 2} e=a_{0}\left(\frac{1}{3}\right)^{2} \)
\(a=\frac{a_{0}}{9} \)

Initially when the sphere in completely filled with water, its centre of gravity (C.G) lies at its centre. As water flows out, the centre of gravity begins to shift below the centre of the sphere. The effective length of the pendulum increases and hence the time period increases. When the sphere is half empty C.G begins to rise up. As the length of pendulum decreases T decreases.

\(\mathrm{T}= \mathrm{T}_{1}=2 \pi \sqrt{\frac{l}{g}} \)
\(T_{1}^{2}=4 \pi^{2} \frac{l}{g} \)
\(\mathrm{y}=\mathrm{kt}^{2} \quad \mathrm{k}=1 \mathrm{~m} \mathrm{~s}^{-2} \)
\(\therefore \mathrm{y}=\mathrm{t}^{2} \quad \mathrm{y} \propto \mathrm{t}^{2} \)
\(\frac{y_{1}}{y_{2}}=\frac{t_{1}^{2}}{t_{2}^{2}} \)
\(\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{y_{1}}{y_{2}}=\frac{6}{5} \)

The sketch between cause (magnitude of acceleration) and effect (magnitude of displacement) is a straight line.

Here,
T = 25 s , At t = 2 s y = 0 and at t = 5 s v = 4 ms^-1,
a =?
For simple harmonic motion, y = a sin wt = a sin \(\frac{2\pi}{2}\)t when t = 4 s, the time taken by particle to travel from the mean position to a given position
= 4 - 2 = 2 s the displacement.
y = a sin\(\left( \frac { 2\pi }{ 16 } \times 2 \right) \) = a sin\(\left( \frac { \pi }{ 4 } \right) =\frac { a }{ \sqrt { 2 } } \)
Velocity v = w\(\sqrt { { a }^{ 2 }-{ y }^{ 2 } } \)
4 =\(\left( \frac { 2\pi }{ 16 } \right) \sqrt { { a }^{ 2 }-\frac { { a }^{ 2 } }{ 2 } } \)
=\(\frac { \pi }{ 8 } \times \frac { a }{ \sqrt { 2 } } \)
a = \(\frac { 32\sqrt { 2 } }{ \pi } \) = 14.4 m

Only incase of resonance, both amplitudes and energy of oscillation are maximum in the conduction of resonance
\({ \omega }_{ 1 }={ \omega }_{ 2 }\)

Yes, every periodic motion need not be simple harmonic motion explain the motion of the earth round the sun is a period motion, but not simple harmonic motion as the back and forth motion is not taking place.

(i) Displacement:
The displacement of a particle executing linear SHM, at an instant is defined as the distance of the particle from the mean position at that instant.
(ii) Velocity:
Is defined as the time rate of change of the displacement of the particle at the given instant.
(iii) Amplitude: The maximum displacement on either side of mean position.
(iv) Acceleration:
It is defined as the time rate of change of the velocity of the particle at the given instant.
(v) Time period:
It is defined as the time taken by the particle executing S.H.M to complete one vibration.

When the oscillator is allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration.

(1) Free oscillation:
(i) The oscillation of a particle with fundamental frequency under the influence of restoring force are defined as free oscillation.
(ii) The amplitude, frequency, and energy of oscillation remains constant.
(iii) Frequency of oscillation is called natural frequency because it depends upon the nature and structure of the body.

(2) Damped oscillation:
(i) The oscillation of a body whos amplitude goes on decreasing with time are defined as damped oscillation.
(ii) In these oscillation the amplitude oscillation decreases exponentially due to damping forces like frictional fore viscouse force, etc.
(iii) Due to decrease in amplitude the energy of the oscillator also goes on decreasing exponentially.

(iv) The force produces a resistance to the oscillation is called damping force.
If the velocity of oscillator is v then Damping force Fd=-bv, b = damping constant.
(v) Resultant force on a damped oscillator is given by
F = F_R+F_d= -kx-kv ⇒ \(\frac { md^{ 2 }x }{ dt^{ 2 } } +b\frac { dx }{ dt } +kx\) = 0
(vi) Displacement of damped oscillator given by
x = \({ x }_{ m }e^{ \frac { bt }{ 2m } }sin(w't+\psi )\)
where w' = angular frequency of the damped oscillator = \(\sqrt { { W }_{ 0 }^{ 2 }-\left( \frac { b }{ 2m } \right) ^{ 2 } } \)
The amplitude decreases continuously with time according to x =\(x_{ m }e^{ \left( \frac { b }{ 2m } \right) t }\)
(vii) For a damped oscillator if the damping is small then the mechanical energy decreases exponentially with time as
E = \(\frac { 1 }{ 2 } kx_{ m }^{ 2 }e^{ \frac { ht }{ m } }\) .

Consider a U -shaped glass tube which consists of two open arms with uniform cross-sectional area A. Let us pour a non-viscous uniform incompressible liquid of density p in the U- shaped tube to a height h as shown in the figure. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O. It means the pressure as meazured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), which balances with the atmospheric pressure. Due to this the level of liquid in each arm will be the same. by blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O which means, the pressure at blown arm is higher than the other arm. This created difference in pressure which will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position and finally comes to rest
Time period of the oscillation is
T = \(2\pi \sqrt { \frac { 1 }{ 2g } } \) second   ......(1)