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 \(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =n\)
\(\frac { { k }_{ 2 } }{ k } =(n+1)\quad \therefore { k }_{ 2 }=k(n+1)\) 

We know that
y = a sin ωt
∴ Velocity of the block = \({dy\over dt}= aω\ cos\ ωt\)
and acceleration of the block \(={d^2y\over dt^2}=-ω^2asin\ ωt=-ω^2y\)
For maximum acceleration y = a
\(∴\ \left(d^2y\over dt^2\right)_{max}=-ω^2a\)
The block will be separated form the piston when
\(ω^2=g\ or\ a={g\over ω^2}\)
or \(s={gT^2\over 4\pi^2}\)
According to the given problem T = 1see
\(a={g\over 4\pi^2}={9.8\over 4\times(3.14)^2}=0.248m/sec^2\)
At this amplitude, the maximum velocity of the block will be
\(ω\ a={2\pi a\over T}={2\times3.14\times0.248\over 1}=1.56m/sec\)

Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount p. so that the ball is depressed to position D, where CD = y.
There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay
Volumetric strain =\(\frac { change\ in\ volume }{ original\ volume } \)
\(=\frac { \Delta V }{ V } =\frac { Ay }{ V } \)
∴Bulk Modulus of elasticity E. will be
\(E=\frac { stress(or\ increase\ in\ pressure) }{ volumetric\ strain } \)
\(=\frac { -p }{ Ay/V } =\frac { -pV }{ Ay } \)
Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.
Now, \(p=\frac { -EAy }{ V } \)
Due to this excess pressure, the restoring force acting on the ball is
\(F=p\times A=\frac { \_ EAy }{ V } .A=\frac { -E{ A }^{ 2 } }{ V } y\quad ...(i)\)
Since F ∝ y and negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, the ball will start executing linear SHM in the neck of chamber with C as mean position.
In S.H.M., the restoring force,
F = -ky ...(ii)
Comparing (i) and (ii), we have
Spring factor, k = EA²/V
Here inertia factor mass of ball = m.
Period, T = \(2\pi \sqrt { \frac { inertia\ factor }{ spring\ factor } } \)
\(=2\pi \sqrt { \frac { m }{ { EA }^{ 2 }/V } } =\frac { 2\pi }{ A } \sqrt { \frac { mV }{ E } } \)
∴ Frequency, v = \(\frac { 1 }{ T } =\frac { A }{ 2\pi } \sqrt { \frac { E }{ mV } } \) 

\(x=2cos\left( 3t+\frac { \pi }{ 3 } +\frac { \pi }{ 2 } \right) \)
Radius of the reference circle, r = amplitude of SHM = 2 cm,
At t = 0, x = - 2sin\(\frac{\pi}{3}=\frac{-2\sqrt3}{2}=-\sqrt3 cm\)
Also ωt = 3t ∴ ω = 3rad/s
\(cos{ \phi }_{ 0 }=-\frac { \sqrt { 3 } }{ 2 } ,{ \phi }_{ 0 }={ 150 }^{ 0 }\)
The reference circle is, thus, as plotted below.

(b) \(x=cos\left( t-\frac { \pi }{ 6 } \right) \)
Radius of circle, r amplitude of SHM = 1 cm
At t= 0, X = cos\(\frac{\pi}{6}=\frac{\sqrt3}{2}cm\)
Also ωt = 1t⇒ ω = 1rad/s
\(cos{ \phi }_{ 0 }=\frac { \sqrt { 3 } }{ 2 } ,{ \phi }_{ 0 }=-\frac{\pi}{6}\)
The reference circle is, thus as plotted below

(c) \(x=3cos\left( 2\pi t+\frac { \pi }{ 4 } +\frac { \pi }{ 2 } \right) \)
Here, radius of reference circle, r = 3 cm and at t = 0, x = 3 sin\(\frac{\pi}{4}=\frac{\sqrt3}{2}cm\)
 ωt = 2\(\pi\)t⇒ ω = 2\(\pi\)rad/s
\(cos{ \phi }_{ 0 }=\frac { \sqrt { \frac { 3 }{ 2 } } }{ 3 } =\frac { 1 }{ \sqrt { 2 } } \)
Therefore,the reference circle is being shown below.

(d) x = 2 cos \(\pi\)t
Radius of reference circle, r = 2 cm and at t = 0, x = 2 cm
∴  ωt = \(\pi\)t or ω = \(\pi\)rad/s
\(cos{ \phi }_{ 0 }=1,{ \phi }_{ 0 }=0\)
The reference circle is plotted below

From Eq, the time period of a simple pendulum is given by,
\(T=2 \pi \sqrt{\frac{l}{g}}\) 
From this relation one gets,
\(l=\frac{g T^{2}}{4 \pi^{2}}\)
The time period of a simple pendulum, which ticks seconds, is 2 s. Therefore, for g = 9.8 m s^–2 and T = 2 s, L is
\(=\frac{9.8(ms^{-2}) \times4(s)^{2}}{4 \pi^2}=1\)
= 1 m

The block executes SHM, its angular frequency, as given by Eq \(ω = \sqrt{\frac{k}{m}}\)
\(ω = \sqrt{\frac{50 Nm^{-1}}{1kg}}\)
= 7.07 rad s^–1
Its displacement at any time t is then given by, x(t) = 0.1 cos (7.07t) Therefore, when the particle is 5 cm away from the mean position, we have 0.05 = 0.1 cos (7.07t)
Or cos (7.07t) = 0.5 and hence sin (7.07t) = \(\frac{\sqrt{3}}{2}\) = 0.866
Then, the velocity of the block at x = 5 cm is = 0.1 × 7.07 × 0.866 m s^–1
= 0.61 m s^–1
Hence the K.E. of the block, \(= \frac{1}{2}\) mv²
= ½[1kg × (0.6123 m s^–1 )² ] = 0.19 J
The P.E. of the block, \(= \frac{1}{2}\) k x² = ½(50 N m^–1 × 0.05 m × 0.05 m) = 0.0625 J
The total energy of the block at x = 5 cm, = K.E. + P.E. = 0.25 J
we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,
= ½(50 N m^–1 × 0.1 m × 0.1 m ) = 0.25 J
which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of conservation of energy.

 As,  E_k = E_p
\(\therefore \)     \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }({ A }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ x }^{ 2 }\)
\(\\ \Rightarrow \quad { A }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\ or\ 2{ x }^{ 2 }={ A }^{ 2 }\)
\(\\ \Rightarrow \quad { x }^{ 2 }=\frac { { A }^{ 2 } }{ 2 } or\ x=\pm \frac { A }{ \sqrt { 2 } } \)
Thus, the energy will be half kinetic and half potential at displacement \(\frac { A }{ \sqrt { 2 } } \) on either side of the mean position.

As, \(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } \)
\(k=4\pi ^{ 2 }m{ v }^{ 2 }\)
For maximum displacement \({ y }_{ max }=A\) 
Maximum restoring force,
F = - kA =- mg
or \(A=\frac { mg }{ k } =\frac { mg }{ 4{ \pi }^{ 2 }{ mv }^{ 2 } } =\frac { g }{ 4\pi ^{ 2 }{ v }^{ 2 } }\)
\( \\ =\frac { 9.8 }{ 4\times { (3.14) }^{ 2 }\times { (0.50) }^{ 2 } } =0.99m\)

Let A be the displacement and \(\omega \)  be the angular frequency of the simple harmonic oscillator.
\(Then,\ \ a_{ 0 }=\omega ^{ 2 }A\ and\ v_{ 0 }=\omega A\)
\(\\ on\ dividing,\ \frac { { v }_{ 0 }^{ 2 } }{ { a }_{ 0 } } =\frac { { \omega }^{ 2 }{ A }^{ 2 } }{ { \omega }^{ 2 }A } =A\ or\ A=\frac { { v }_{ 0 }^{ 2 } }{ { a }_{ 0 } } \)

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.

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.

As value of g on moon is less than that on earth, in accordance with the realtion T =2π\( \sqrt{l/g}\), the time period of oscillations of a simple pendulum on moon will be greater.

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

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
\(\therefore \frac { { v }_{ 1 }^{ 2 } }{ { v }^{ 2 } } =\frac { { g }_{ 1 } }{ g } =\frac { \frac { 3g }{ 2 } }{ g } =\frac { 3 }{ 2 } \)
\(or\quad \frac { { v }_{ 1 } }{ v } =\sqrt { \frac { 3 }{ 2 } } =1.225\)
\(\Rightarrow { v }_{ 1 }=1.225v=1.225\times \frac { 1 }{ 2 } =0.612{ s }^{ -1 }\) 

For 1st particle
y₁= 4 sin (10t +θ)
Velocity = \(\frac{dy_1}{dt}\)=4 \(\times\) 10 cas (10t +θ)
For second particle
y₂= 5 cos 10t = 5 sin(10t + \(\pi/2\))
Velocity= \(\frac{dy_2}{dt}\) = 5 \(\times\)10 cos (10t + \(\pi/2\))
= 5 sin(10t+\(\pi/2\))
Phase difference between velocities
=(10t + θ)-(10 t+ \(\pi/2\))
=\(\left( \theta -\frac { \pi }{ 2 } \right) \) 

If in a spring mass system, the mass is displaced and given a velocity, also it will perform SHM but with an amplitude more than the amount of extension.
We have for SHM
x = A cos\((\omega t+\theta )\)
where,x = displacement, A = amplitude
\(\theta \) = phase constant, we get
On differenting with respect to t,
\(\frac { dx }{ dt } =-A\omega sin(\omega t+\theta )\)
\(\\ \Rightarrow \ v=-A\omega sin(\omega t+\theta \)
where, v= instantaneous velocity of the particle at t.
Now , at t = 0,x = x₀
From Eq(i)
\(\Rightarrow { x }_{ 0 }=Acos\theta \)
Again at t = 0, v = v₀
FromEq(ii)
\( -{ v }_{ 0 }=-A\omega sin\theta \ or\ \frac { { v }_{ 0 } }{ \omega } =Asin\theta \)
\(\\ { A }^{ 2 }({ cos }^{ 2 }\theta +{ sin }^{ 2 }\theta )={ x }_{ 0 }^{ 2 }+\left( \frac { v_{ 0 } }{ \omega } \right) ^{ 2 }\)
\( \Rightarrow A^{ 2 }={ x }_{ 0 }^{ 2 }+\left( \frac { v_{ 0 } }{ \omega } \right) ^{ 2 }\ \)
\(\\ or A=\sqrt { { x }_{ 0 }^{ 2 }+\left( \frac { { v }_{ 0 } }{ \omega } \right) ^{ 2 } } \)