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

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.

(i) The motion is repeated after a certain interval of time, hence periodic. In fact, the bar magnet oscillates about its mean position with a definite period of time.
(ii) Rorarary motion is periodic as repeating after fixed time-interval.
(iii) There is no repetition, hence not periodic.

For SHM, the restoring force on the particle must be proportional to its displacement and directed towards mean position.

Time period, \(T=2 \pi \sqrt{\frac{l}{g}}\) or \(T \alpha \sqrt{l}\)
% increase in time period
\(\frac{\Delta T}{T} \times 100=\frac{1}{2} . \frac{\Delta l}{l} \times 100=\frac{1}{2} \times 25=12.5 \%\)

According to force equation of SHM, F = - kx, where k is a constant known as force constant.

−0.4 m/s²

(i) - 6.0cm
(ii) 16 Hz

Here I = l₁ + l₂ ...(i)
and l₁ = nl₂
We know \(k=\frac { Mg }{ l } ...(iii)\)
\(k_{ 1 }=\frac { Mg }{ l_{ 1 } } ...(iv)\)
and \(k_{ 2 }=\frac { Mg }{ l_{ 2 } } ...(v)\)
Dividing equation (iv) by equation (iii) we find 
\(\frac { { k }_{ 1 } }{ k } =\frac { l }{ { l }_{ 1 } } =\frac { { l }_{ 1 }+{ l }_{ 2 } }{ { l }_{ 1 } } =1+\frac { { l }_{ 2 } }{ { l }_{ 1 } } \)
From equation (ii), we find \(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =n\)
\(\frac { { k }_{ 1 } }{ k } =1+\frac { 1 }{ n } or{ k }_{ 1 }=\left( \frac { n+1 }{ n } \right) k\)
From equation (v) and (iii), we find:
\(\frac { { k }_{21 } }{ k } =\frac { l }{ { l }_{ 2 } } =\frac { { l }_{ 1 }+{ l }_{ 2 } }{ { l }_{ 1 } } =\frac { { l }_{1 } }{ { l }_{ 2 } }+1 \)
From equation (ii) we have \(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =n\)
\(\frac { { k }_{ 2 } }{ k } =(n+1)\quad \therefore { k }_{ 2 }=k(n+1)\) 

Two simple harmonic motions of equal amplitudes (A) and frequencies ω and 2ω and initial phase difference of π/2 are represented by
x = A sin ωt     ...(i)
\(y=A\ sin\left({2ω+{\pi\over 2}}\right)=A\ cos\ 2 \omega t\)       ...(ii)
Since cos 2 ωt (1 - 2 sin² ωt)
∴ y = A[1 - 2 sin² ωt]      ...(iii)
From eqn. (i) 
\(sin^2\ ωt={x^2\over A^2}\)
\(∴\ \ y=A\left[1-{2x^2\over A^2}\right]=A-{2x^2\over A}\)
\(⇒\ \ {2x^2\over A}+y-A=0\)
or \(x^2+{Ay\over 2P}-{A^2\over 2}=0\)
which is the equation of a parabola. Hence the resultant path followed by the particle is parabolic.

(a) Let y be the maximum extension produced in the spring in Fig. (a)
Then F = ky (in magnitude) ∴ y = \(\frac{F}{k}\)
If fig. (b), the force on one mass acts as the force of reaction due to the force on the other mass. Therefore, each mass behaves as if it is fixed with respect to the other.
Therefore, F = ky ⇒ y= \(\frac{F}{k}\)
(b) In fig. (a), F = -ky
⇒ ma = -ky ⇒  a = - \(\frac{k}{m}y\) ∴ω²=  \(\frac{k}{m}\) i.e.,ω =\(\sqrt{\frac{k}{m}}\)
Therefore, period \(T=\frac { 2\pi }{ \omega } =2\pi \sqrt { \frac { m }{ k } } \)
In fig. (b), we may consider that the centre of the system is 0 and there are two springs each of length \(\frac{l}{2}\) attached to the two masses, each m, so that k' is the spring factor of each of the springs.
Then, K' = 2k

∴ \(T=2\pi \sqrt { \frac { m }{ k' } } \)
\(T=2\pi \sqrt { \frac { m }{ 2k } } \) 

Given equation, \(x=12\sin { \left( 10t+0.6 \right) } \) 
On comparing with \(x\left( t \right) =A\sin { \left( \omega t+\phi \right) } \) 
We have, 
(i) Amplitude, A = 12 m
(ii) Angular frequency,\(\omega =10rad/s\) 
(iii) Frequency, \(v=\frac { \omega }{ 2\pi } =\frac { 10 }{ 2\pi } =1.59Hz\) 
(iv) Time period, \(T=\frac { 2\pi }{ \omega } =\frac { 1 }{ 1.59 } =0.628s\)  
(v) Initial phase, \(\omega t+{ \phi }|_{ t=0 }={ 10t+0.6 }|_{ t=0 }\)  
\(=0.6rad\).

(i) Boys were sensible and were true to their words.
(ii)  \(F=20\ N,\ \Delta l=0.2m,\ m\ =\ 4kg\)
\( (a)\ k=\frac { F }{ \Delta l } =\frac { 20 }{ 0.2 } =100{ Nm }^{ -1 }\)
\( (b)\ y=\frac { mg }{ k } =\frac { 4\times 10 }{ 100 } =0.4\ m\)
(iii) Watch, winding keys of toys.

The object will not detach from the platform, if the angular frequency ω is such that, during the downward motion, the maximum acceleration equals the acceleration due to gravity, i.e
\(ω_{max}^2A=g\)
or \(ω_{max}=\sqrt{g\over A}\)
or \(T_{min}={2\pi\over ω_{max}}=2\pi\sqrt{A\over g}\)
Now A 2.5 cm = 2.5 x 10^-2 m and g = 10 ms=^-2.
Substituting these values we get
\(T_{min}={\pi\over 10}=0.314s\)

(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 \(T=2\pi \sqrt { \frac { m }{ k } } =2\times 3.14\times \sqrt { \frac { 3 }{ 147 } } =0.9s\) 

In this case, the bob of the pendulum is under the action of two accelerations.
(i) Acceleration due to gravity 'g' acting vertically downwards.
(ii) Centripetal acceleration a_c= \(\frac{v^2}{R}\) acting along the horizontal direction.
∴ Effective acceleration, g'= \(\sqrt { { g }^{ 2 }+{ a }_{ c }^{ 2 } } \)
\(or \ g'=\sqrt { { g }^{ 2 }+\frac { { v }^{ 4 } }{ { R }^{ 2 } } } \)
Now time period,\(T'=2\pi \sqrt { \frac { 1 }{ g } } =2\pi \sqrt { \frac { 1 }{ \sqrt { { g }^{ 2 }+\frac { { v }^{ 4 } }{ { R }^{ 2 } } } } } \)  

A second pendulum means a simple pendulum having time period T = 2s
For a simple pendulum T\(=2\pi \sqrt { \frac { l }{ g } } \)
where, l = length of the pendulum and g= acceleratin due to gravity On surface of the earth.
\({ T }_{ s }=2\pi \sqrt { \frac { l_{ e } }{ { g }_{ e } } } \)
On the surface of the moon,
\({ T }_{ n }=2\pi \sqrt { \frac { l_{ m } }{ { g }_{ m } } } \)
Dividing Eq(i) by Eq(ii) we get
\(\frac { { T }_{ s } }{ { T }_{ m } } =\frac { 2\pi }{ 2\pi } \sqrt { \frac { l_{ e } }{ { g }_{ e } } } \times \sqrt { \frac { l_{ m } }{ { g }_{ m } } } \)
T_s = T_m to maintain the second pendulum time period.
\(1=\sqrt { \frac { l_{ e } }{ { g }_{ e } } \times \frac { l_{ m } }{ { g }_{ m } } } \)
But the acceleration due to gravity at moon is 1/6 of the acceleration due to gravity at earth i.e.\({ g }_{ m }=\frac { { g }_{ e } }{ 6 } \) 
Squaring Eq (iii) and putting this value
\( 1=\frac { { l }_{ e } }{ { l }_{ m } } \times \frac { { g }_{ e }/6 }{ { g }_{ e } } =\frac { { l }_{ e } }{ { l }_{ m } } \times \frac { 1 }{ 6 } \)
\(\\ \Rightarrow \frac { { l }_{ e } }{ { 6l }_{ m } } =1\)
\(or { l }_{ m }=\frac { 1 }{ 6 } { l }_{ e }=\frac { 1 }{ 6 } \times 1=\frac { 1 }{ 6 } m\)