In first case
T₁=\(2\pi \sqrt { \frac { l }{ g } } \)
in second case, displacement y = kt2
upward velocity, v =\(\frac{dy}{dt}\)=2 kt.
upward acceleration a = 2,
k = 2\(\times\)1 ms^-2 = 2 ms^-2
T₂= \(2\pi \sqrt { \frac { l }{ g+a } } =2\pi \sqrt { \frac { l }{ g+2 } } \)
Hence \(\frac { { T }_{ 1 }^{ 2 } }{ { T }_{ 2 }^{ 2 } } =\frac { 4{ \pi }^{ 2 }l }{ g } \times \frac { g+2 }{ 4{ \pi }^{ 2 }l } \)
=\(\frac { g+2 }{ g } =\frac { 10+2 }{ 10 } =\frac { 6 }{ 5 } \)

Velocity of particle 1
V₁=\(\frac { d{ y }_{ 1 } }{ dt } =0.1cos(100\pi t+\frac { \pi }{ 3 } )\times 100\pi \)
=10\(\pi\) cos(100 \(\pi\)t+\(\frac{​​\pi }{3}\))
Velocity of particle 2
V₂=\(\frac { d{ y }_{ 2 } }{ dt } =0.1(-sin\pi t)\times p\)
= -0.1\(\pi\) sin\(\pi\)t
= 0.1 cos(\(\pi\)t+\(\frac{\pi}{2}\))
Phase difference of the velocity of particle 1 With respect to the elecity of particle 2 is
\(\triangle \phi ={ \phi }_{ 1 }-{ \phi }_{ 2 }=\frac { \pi }{ 3 } -\frac { \pi }{ 2 } =-\frac { \pi }{ 6 } \)

Let \(\theta\) be the angle made by the string with vertical. When the bob is at the mean position \(\theta\) = 0.

(ii) These are two forces acting on the bob the tension T along the vertical force due to gravity (= mg).
(iii) The force can be resolved into mg cos \(\theta\) along a circle of length L and centre at this support point.
(iv) Its radial acceleration (w²L) and also tangential acceleration. So net radial force = T - mg cos \(\theta\), while the tangential acceleration provided by mg sin \(\theta\). Since the radial force gives zero torque.
(v) So torque provided by the tangential component.
\(\tau\) = Lmg sin\(\theta\)
\(\tau\) = \(\alpha\)(by Newton's law of rotational motion)
I\(\tau\) =mg sin\(\theta\) L
\(\alpha =\frac { mgL }{ I } sin\theta \)
Where I is the moment of inertia, a \(\alpha\) angular acceleration,
(iv) The \(\theta\) is to small,
sin \(\theta\) = \(\theta\)
\(\alpha =-\frac { mgL }{ I } \theta \)
\(w=\sqrt { \frac { I }{ mgL } } \)
[for simple harmonic a (t) = -w²t \(\times\) \(\theta\) is small] and T =2\(\pi \sqrt { \frac { I }{ mgL } } \) \(\left[ \therefore w=\frac { 2\pi }{ T } \right] \)

(i) The suction pump creates the pressure difference. Hence mercury rises in one limb of the U-tube.
(ii) When it is removed a net force acts on the liquid column due to the difference in level of mercury in the two limbs and therefore the liquid column executes S.H.M. Which can be explained as follows.

(iii) The mercury contained in a vertical U-tube upto the level A and B in its two limbs.
Suppose \(\rho\) = density of the mercury.
Where, L is total length of the mercury column in both the limbs.
A is internal cross-sectional area of U-tube
M is mass of mercury in U-tube = L_Ap
(iv) Suppose the mercury be depressed in left links to A' by the small distance y, then it rises by the same amount in the right limb to position B'.
∴ Difference in levels in the two limbs.
= A'B' =2y
(v) Volume of mercury contained in the column of length
2y = A\(\times\)2y
m = A\(\times\)2y\(\times\) \(\rho\)
(vi) When w = weight of liquid contained in the column of length 2y
Then w = mg = t\(\times\)2y\(\times\)\(\rho\) \(\times\)g
This weight produced the restoring force F which tends to bring back the mercury to its equilibrium position.
F = -2 Ay\(\rho\)p = -(2Apg)y
(vii) When a = acceleration produced in the liquid column, then
a =\(\frac { F }{ m } =\frac { (2A\rho g)y }{ LAP } \)
or a =\(\frac { -2\rho yg }{ L\rho } =\frac { 2\rho g }{ 2Lp } y\)    ........(1)
[∴ L=2h]
Where, h is height of mercury in each limb. Now from equation (1), if is clear that a ay and Negative sigh shows that it acts opposite to y, so the motion of mercury in U-tube is simple harmonic in nature having time period (T) given by T =\(2\pi \sqrt { \frac { y }{ a } } =2\pi \sqrt { \frac { 2h\rho }{ 2\rho g } } \)
= \(2\pi \sqrt { \frac { h\rho }{ \rho g } } \)
T = \(2\pi \sqrt { \frac { h }{ g } } \) .

(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 = \(\frac { 1 }{ 2 } kx_{ m }^{ 2 }e^{ \frac { ht }{ m } }\) .