Let the initial mass and time periods be M and T respectively. If Hooke's law is obeyed, then the oscillations of the spring will be simple harmonic having time period T given by
\(T=2\pi \sqrt { \frac { M }{ k } } \)
Given T = 2 s
\(2=2\pi \sqrt { \frac { M }{ k } } \)k = spring constant...(i)
On increasing the mass by 2 kg
\(3=2\pi \sqrt { \frac { M+2 }{ k } } \)....(ii)
Squaring and dividing equation (ii) by (i) we have
\(\frac { 9 }{ 4 } =\frac { M+2 }{ M } =1+\frac { 2 }{ M } \)
\(or\quad \frac { 2 }{ M } \quad =\frac { 9 }{ 4 } -1=\frac { 5 }{ 4 } \)
\(\therefore M=\frac { 2\times 4 }{ 5 } =\frac { 8 }{ 5 } =1.6kg\) 

For second pendulum, frequency v = \(\frac{1}{2}\)s^-1
When elevator is moving upwards with acceleration a, the effective acceleration due to gravity is g₁= g + a = g + \(\frac{g}{2}\)=\(\frac{3g}{2}\)
Since,\(v=\frac { 1 }{ 2\pi } \sqrt { \frac { g }{ l } } \)
Hence, v² ∝g
\(\therefore \frac { { v }_{ 1 }^{ 2 } }{ { v }^{ 2 } } =\frac { { g }_{ 1 } }{ g } =\frac { \frac { 3g }{ 2 } }{ g } =\frac { 3 }{ 2 } \)
\(or\quad \frac { { v }_{ 1 } }{ v } =\sqrt { \frac { 3 }{ 2 } } =1.225\)
\(\Rightarrow { v }_{ 1 }=1.225v=1.225\times \frac { 1 }{ 2 } =0.612{ s }^{ -1 }\) 

Area of cross-section of the block = \(\pi\)r² = 15 cm²
= 15 x 10^-4 m²
Total weight of the block = (230 + 50) = 280 gm = 0.28 kg
Density of wood = 0.30 gm/ c.c. = 300 kg/ m³
Density of water = 103 kg/ m³
When the cylinder is depressed in water through a distance y, the Restoring force weight of water displaced
F = Aydg = (15 x 10^-4) x 10³ x 9.8 Newton
Restoring force per unit distance= \(\frac{F}{y}\) = k
= (15 x 10^-4) x 10³ x 9.8 newton/ metre
= 1.5 x 9.8 N/m
Hence the frequency of oscillation is given by
\(=\frac { 1 }{ 2\pi } \sqrt { \left( \frac { k }{ m } \right) } =\frac { 1 }{ 2\pi } \sqrt { \frac { 1.5\times 9.8 }{ 0.28 } } \)
= 1.15 Hz

(i) and (ii) since acceleration of the lift is zero therefore there will be no effect on time period.
(iii) When the lift moves up with uniform acceleration at the effective value of acceleration due to gravity is g + a.
\(\therefore T'=2\pi \sqrt { \frac { l }{ g+a } } \)Clearly,T'
(iv) When the lift moves down with uniform acceleration a, then the effective value of g is g - a.
\(\therefore T'=2\pi \sqrt { \frac { l }{ g-a } } \)Clearly,T'
(v) When the lift begins to fall freely under gravity, the effective vlaue of g becomes zero. So, T is infinite i.e., the simple pendulum shall not oscillate.

The situation is shown in the fig. When the mass is displaced along the direction of the length of the spring, one spring is compressed while the other is extended but the force due to both the springs is in the same direction. Hence the effective force constant

k = k₁ + k₂ = 8 N/m + 2 N/m = 10 N/m
The frequency of vibration is given by
\(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } =\frac { 1 }{ 2\pi } \sqrt { \frac { 10 }{ 0.1 } } \)
\(or\quad v=\frac { 10 }{ 2\pi } =\frac { 5 }{ \pi } { s }^{ -1 }\) 

Let x = A cos (ωt + \(\phi\)) and if phase angle \(\phi\) is zero, then
x = A cos ωt
\(\therefore \quad v=\frac { dx }{ dt } =-A\omega \quad sin\omega t\)
\(and\quad a=\frac { dv }{ dt } =-a{ \omega }^{ 2 }cos\omega t=-{ \omega }^{ 2 }x\)
Thus, values of z, v and a at different times, over one complete oscillation cycle,

With the given data we plot x-t, v-t and a-t graphs. The graphs have been shown in Fig. (a), (b) and (c).

(a) Force constant \(k=\frac { F }{ l } =\frac { mg }{ l } \)
Here m = 3.0 kg and elongation in length of spring l = 0.2 m
∴ Force constant k = \(\frac{3.0\times 9.8}{0.2}=174Nm^{-1}\)
(b) Period of oscillation \(T=2\pi \sqrt { \frac { m }{ k } } =2\times 3.14\times \sqrt { \frac { 3 }{ 147 } } =0.9s\) 

\(T=2\pi \sqrt{\frac{l}{g}}, l_1=100 cm\ ; l_2=110.25 cm\)
For smaller pendulum, \(T_1=2\pi \sqrt{\frac{100}{g}}\)    -- (i)
For larger pendulum, \(T_1=2\pi \sqrt{\frac{110.25}{g}}\)   -- (ii)
Let these pendulums oscillate in phase again if larger pendulum completes' n' oscillations. It means smaller pendulum must complete (n + 1) oscillations.
\(nT_2=(n+1)T_1\)
or \(\frac{n+1}{n}=\frac{T_2}{T_1}=\sqrt{\frac{110.25}{100}}=1.05\)
or \(1+\frac{1}{n}= 1.05 \ or \ \frac{1}{n}=0.05 = \frac{5}{100}=\frac{1}{20}\)
∴   n = 20.
Hence both pendulums will again oscillate in phase after 20 oscillations of the larger or 21 oscillations of the smaller pendulum.

We know that,\(v=\omega \sqrt { \left( { a }^{ 2 }-{ y }^{ 2 } \right) } \)
Further \(\omega =\frac { 2\pi }{ T } \)
\(v=\frac { 2\pi }{ T } \sqrt { ({ a }^{ 2 }-{ y }^{ 2 }) } =\frac { 2\times 3.14 }{ \left( \frac { 11 }{ 7 } \right) } \sqrt { [({ 1.0) }^{ 2 }-({ 0.6) }^{ 2 })] } \)
= 3.2 m/sec.
Kinetic energy at this displacement is given by
\(K=\frac { 1 }{ 2 } { mv }^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 0.8\times { (3.2) }^{ 2 }=4.1joule\) 

This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by
\(T=2\pi \sqrt { \frac { l }{ g } } i.e,\quad T\infty \sqrt { l } \)
When the girl stands up, the distance between the point of suspension and the centre of mass of the swinging body decreases i.e.,l decreases, so T will also decrease.

\(Here,m=10g,T=6s\)
\(\omega =\frac { 2\pi }{ T } =\frac { 2\pi }{ 6 } =\frac { \pi }{ 3 } rads^{ -1 }\)
\( When\ t=1s,v=6cms^{ -1 }\)
\(\\ As\ v=A\omega cos\omega t\)
\( 6=A\times \frac { \pi }{ 3 } cos\frac { \pi }{ 3 } \times 1=A\times \frac { \pi }{ 3 } cos{ 60 }^{ \circ }\)
\(=A\times \frac { \pi }{ 3 } \times \frac { 1 }{ 2 } =\frac { \pi A }{ 6 } orA=\frac { 36 }{ \pi } cm\)
\(Total\ energy,E=\frac { 1 }{ 2 } m{ A }^{ 2 }\omega ^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 10\times \left( \frac { 36 }{ \pi } \right) ^{ 2 }\times \left( \frac { \pi }{ 3 } \right) ^{ 2 }=720erg\)
\(\\ Kinetic\ energy=\frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } \times 10\times 6^{ 2 }=180erg\)
\(\\ \therefore Potential\ energy=Total\ energy-Kinetic\ energy\)
\( =720-180=540erg\)

\(When\ t=0,x=+A\)
\(\\ x(t)=Asin(\omega t+\phi )\ at\ t=0\ and\ x=+A\)
\(+A=Asin(\omega \times 0+\phi )\)
\( or\ 1=sin\phi \Rightarrow \phi =\frac { \pi }{ 2 } \)
\(\\ \therefore \ x(t)=Asin\left( \omega t+\frac { \pi }{ 2 } \right)\)
\(=Acos\omega t=Acos20t=2cos20t\)

If we replace \(sin\theta \approx \theta \) for large angles, then actually \(sin\theta <\theta \)
Now since this factor is multiplied to the restoring force mg\(sin\theta \) is replaced by \(mg\theta \) which means ana effective reduction in g for large angles. hence, there is an increase in time period T over that given by the formula  \(T=2\pi \sqrt { \frac { l }{ g } } \) as compared to the acse which it is assumed \(sin\theta \simeq \theta \)               

\(T=0.314\ s,\ { v }_{ max }=1{ ms }^{ -1 },\ E=1.5\ J\)