\(\frac{1}{2}\)mω²r² where r = amplitude, ω = angular frequency, m = mass of the oscillator.

When amplitude of the vibrating pendulum is small, then angular displacement of the bob used in simple pendulum is small. Here the restoring force F = mg sin θ = mgθ = mgx/l, where x is the displacement of the bob and I is the length of pendulum. Hence F ex: x. Since F is directed towards mean position, therefore the motion of the bob of simple pendulum will be S.H.M. if θ is small

Force constant k of parallel combination is given by k = k₁+ k₂. Thus, force constant of the parallel combination is equal to the sum of individual force constants of two springs.

No change, since \(T=2\pi \sqrt { \frac { m }{ k } } \)

\(\frac { K.E }{ Total\ energy } =\frac { \frac { 1 }{ 2 } { m\omega }^{ 2 }\left( { a }^{ 2 }-\frac { { a }^{ 2 } }{ 4 } \right) }{ \frac { 1 }{ 2 } { m\omega }^{ 2 }{ a }^{ 2 } } =\frac { 3 }{ 4 } \)

This is because of resonant vibrations.

\(T=2\pi \sqrt { \frac { l }{ g } } \)so it does not depend upon the mass.

Yes, it is if the spring is perfectly elastic.

No. This is because of zero value of g at the centre of Earth.

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

Given  PE = KE
i.e \(\frac { 1 }{ 2 } m\omega ^{ 2 }{ x }^{ 2 }=\frac { 1 }{ 2 } { m\omega }^{ 2 }({ A }^{ 2 }-{ x }^{ 2 })\)
\({ x }^{ 2 }={ A }^{ 2 }-{ x }^{ 2 }\Rightarrow x=\frac { A }{ \sqrt { 2 } } \)
\(\\ Now\ x=A\ sin\ \omega t=A\ sin\left( \frac { 2\pi }{ T } \right) t\)
\(\\ So,\ \frac { A }{ \sqrt { 2 } } =A\ sin2\pi \frac { t }{ 8 } \)
\(\\ or\ sin\frac { \pi t }{ 4 } =\frac { A }{ \sqrt { 2 } } =sin\frac { \pi }{ 4 } \)
\(\\ or\ \frac { \pi t }{ 4 } =\frac { \pi }{ 4 } \ or\ t=1s\)

A force which takes the body towards the mean postion in oscillation is called restoring force.
Characterstic of Restoring Force
The restoring force is aleays directed towards the mean positin and its magnitude of any instant is directly. Proportional to the displacement of the particle froom its mean postion of that instance.

Given mass of the disc m = 10 kg
Radius of the disc r = 15 cm = 0.15 m
T = 1,5 s
I is the moment of inertia of the disc about the axis of rotation which is perpendicular to the plane of the disc and passing through its centre.
 \(I=\frac { 1 }{ 2 } { mr }^{ 2 }=\frac { 1 }{ 2 } \times (10)\times (0.15)^{ 2 }\)
= 0.1125 kg-m²
Time period, \(T=2\pi \sqrt { \frac { 1 }{ \alpha } } \)
\(\alpha =\frac { { 4\pi }^{ 2 }I }{ T^{ 2 } } =\frac { 4\times ({ 3.14) }^{ 2 }\times 0.1125 }{ { (1.5) }^{ 2 } } \)
\(\\ =1.972\ Nm/rad\)

(i) Here F = 10 N,\(\triangle l=0.1m,m=4kg\) 
\(k=\frac { F }{ \triangle l } =\frac { 10 }{ 0.1 } =100Nm^{ -1 }\)
(ii) Here F = 10 N,\(\triangle l=0.1m,m=4kg\)
\(y=\frac { mg }{ k } =\frac { 4\times 10 }{ 100 } =0.4m\)
(iii) Here F = 10 N,△l = 0.1m,m = 4kg
\(T=2\pi \sqrt { \frac { m }{ k } } =2\times \frac { 22 }{ 7 } \sqrt { \frac { 4 }{ 100 } } =1.26s\)
(iv) Here F = 10 N,△l = 0.1m,m = 4kg
Frequency, \(v=\frac { 1 }{ T } =\frac { 1 }{ 1.26 } =0.8\quad Hz\)          

Given m = 12Kg, original length l = 50 cm
K = 2.0 x 10³ Nm^-1
F = ky
\(y=\frac { F }{ k } =\frac { mg }{ k } =\frac { 12\times 9.8 }{ 2\times 10^{ 3 } } =5.9\times 10^{ -2 }m=5.9\quad cm\)
Stretched length of the spring = l + y = 50 + 55.9 cm
= 105.9 cm
Frequency of oscillations, \(v=\frac { 1 }{ T } =\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } \)
\(=\frac { 1 }{ 2\times 3.14 } \sqrt { \frac { { 2\times 10 }^{ 3 } }{ 12 } } =2.06{ s }^{ -1 }\)