Figure shows a tunnel dug along the diameter of the earth. Consider the case of a particle of mass m at a distance y from the centre of the earth. There will be a gravitational attraction of the earth on this particle due to the portion of matter contained in a sphere of radius y. The mass of the sphere of radius y is given by

M=Volume x density
or M=\(\frac { 4 }{ 3 } \pi { y }^{ 3 }\times d\)
(where d = density of earth).
This mass can be regarded as concentrated at the centre of the earth. The force F between this mass and the particle of mass m is given by
\(F=-\frac { GMm }{ { y }^{ 2 } } \)
Negative sign shows that the force is of attraction.
\(\therefore F=-G\left( \frac { 4 }{ 3 } \pi { y }^{ 3 }d \right) \frac { m }{ { y }^{ 2 } } =-G\times \left( \frac { 4 }{ 3 } \pi md \right) y\)
or F∝y.
The force is directly proportional to the displacement, hence the motion is simple harmonic motion.
Here ,the constant k=\(\frac { 4 }{ 3 } \pi mdG\)
The time period, T=\(2\pi \sqrt { (m/k) } \)
\(=2\pi \sqrt { \left( \frac { 3m }{ 4\pi mdG } \right) } =2\pi \sqrt { \left( \frac { 3 }{ 4\pi dG } \right) } \)
\(=\sqrt { \left( \frac { 3\pi }{ dG } \right) } =\sqrt { \left( \frac { 3\times 3.14 }{ 5.51\times { 10 }^{ 3 }\times 6.67\times { 10 }^{ -11 } } \right) } \)
= 42.2 minutes.

The particle velocity in SHM is given by : v = ω\(\sqrt{A^2-x^2}\)where A is the amplitude of oscillation.
For displacement x = x₁, v₁ =\(ω\sqrt{A^2-x_1^2}\)or \(v_1^2=ω^2(A^2-x_1^2)\)
and for displacement x = x₂' \(v_2=ω\sqrt{A^2-x_2^2}\) or \(v_2^2=ω^2(A^2-x_2^2)\)
Subtracting (i) from (ii), we have
\(v_2^2=v_2^2=ω^2\left(x_1^2-x_2^2\right)\)
∴ Period of oscillation \(T={2\pi\over ω}={2\pi}\sqrt{(x_1^2-x_2^2)\over (v_2^2-v_1^2)}\)

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.

Suppose by falling down through a height h, the mass m compresses the spring balance by a length x.
This P.E. lost by the mass = mg (h + x)
This is stored up as energy of the spring by compression
\(={1\over 2}kx^2\)
∴ mg (h + x) = \({1\over 2}kx^2\) or \({1\over 2}kx^2-mgx - mgh = a\)
or \(x^2-{2mgx\over k}-{2mgh\over k}=0\)
Solving this quadratic equation, we get
\(x={{2mg\over k}\pm\sqrt{\left(2mgh\over k\right)^2+\left(8mgh\over k\right)}\over 2}\)
In the equilibrium position, the spring will be compressed through the distance mg / k and hence the amplitude oscillation is
\(A={mg\over k}\sqrt{1+{2kh\over mg}}\)
Energy of oscillation \(={1\over 2}kA^2={1\over 2}k\left(mg\over k\right)^2\left(1+{2kh\over mg}\right)\)
\(=mgh+{(mg)^2\over 2k}\)

\(F=-{dU\over dx}=-{d\over dx}[4(1-cos2x)]\)
= - 8 sin 2x
or F = - 8 x 2x
or F = -16 x
As F ∝ x and - ve sign shows that x is directed towards equilibrium position, hence the  particle will execute SHM.
Here, spring factor, k = 16 N/m
inertia factor, m 1 kg
∴ time period, \(T={2\pi}\sqrt{m\over k}=2\pi\sqrt{1\over 16}={\pi\over 2}s\)

Let us suppose that an extension x is produced in the spring when a force mg is applied to it. The equilibrium position in case (a) is given by
F = mg = kx
The time period in this case is given by
\(T_a=2\pi\sqrt{mass\over spring\ constant}=2\pi\sqrt{m\over k}\)
(b) In this case, the length of the spring is doubled. Hence a given force II1g will double the extension. Let x' be the extension produced and k_eff be the force constant of the combination. Thus
F = mg = k_eff x'
= 2 k_eff x (∵ x' = 2x)
Comparing (1) and (3) we get
F = k x = 2 k_effx
or \(k_{eff}={k\over 2}\)
Thus the time period in this case is given by 
\(T_b=2\pi\sqrt{m\over k_{eff}}=2\pi\sqrt{2m\over k}=\sqrt{2}T_a\)
(c) In this case, the extension x" produced in each spring by a force mg is half that produced in case (a), i.e.,
\(x''={x\over 2}\)
If k_eff is the force cosntant of the combination in this case, we have
F = mg = k_eff  x"\(={k_{eff}\over 2}x\)
Comparing (5) with (1) we have
k_eff = 2k
Hence, \(T_e=2\pi\sqrt{m\over k_{eff}}=2\pi\sqrt{m\over k}={T_a\over \sqrt2}\)
From (2), (4) and (6) we have
\(T_a:T_b:T_c=\sqrt2:{1\over \sqrt2}\)

Consider a body of mass M suspended by two springs connected in parallel as shown in Fig. (n) Let k₁ and k₂ be the spring constants of two springs respectively. Let the body be pulled down so that each spring is stretched through a distance y. Restoring forces F₁ and F₂ will be developed in the springs S₁ and S₂ respectively.

According to Hooke's law, F₁ = - k₁y
and F₂ = -k₂y
Since both the forces acting in the same direction, therefore, total restoring force acting on the body is given by
F = F₁ + F2 = - k₁ y - k₂ Y = - (k₁ + k₂) y
∴ Acceleration produced in the body is given by 
\(a={F\over M}=-{(k_1+k_2)y\over M}\)    ..(i)
since \({(k_1+k_2)\over M}\) is constant ∴ a ∝ - y
Hence motion of the body is SHM.
Time period of body is given by
\(T=2\pi\sqrt{y\over |a|}=2\pi\sqrt{M\over k_1+k_2}\ \ ...(ii)\)
If  k₁=k₂=k
Then \(T=2\pi\sqrt{M\over 2K}\)
For series: Consider a body of mass M suspended by two springs S₁ and S₂ which are connected in series as shown in Fig. (b). Let k₁ and k₂ be the spring constants of springs S₁ and S₂ respectively. Suppose at any instant, the displacement of the body from equilibrium position is y in the downward direction. If y₁ and y₂ be the extension produced in the springs S₁ and S₂ respectively, then

y = y₁ + y₂   ...(i)
Restoring forces developed in S₁ and S₂ are given by
F₁ = - k₁ + y₁    ....(ii)
F₂ = - k₂ + Y₂     ....(iii)
Multiplying eqns. (ii) by k₂ and eqn. (iii) by k₁ and adding, we get
∴   k₂ F₁ + k₁ F₂ = - k₁ k₂ (y_l + y₂) = - k₁ k₂ y
[From eqn. (i)]
Since both the springs are connected in series, so
F₁ = F₂ = F
∴ F(k₁ + k₂) = -k₁k₂y or \(F=-{k_1k_2\over (k_1+k_2)y}\)
If a be the acceleration produced in the body of mass M, then
\(a={F\over M}=-{k_1k_2y\over (k_1+k_2)M}\)     ...(ii)
Time period of the body is given by
\(T=2\pi\sqrt{y\over |a|}=2\pi\sqrt{(k_1+k_2)M\over k_1k_2}\)
\(T=2\pi\sqrt{\left({1\over k_1}+{1\over k_2}\right)M}\)

(1) Let A be any point on the circle of reference of the fig. (a) From A, draw BN perpendicular on x-axis.
If ∠POA = θ then
∠OAM = θ = ωt
∴ In triangle OAM
\(\frac{OM}{OA}\) = sinθ

∴ \(\frac { -x }{ 3 } =sin\omega t=sin\frac { 2\pi }{ T } t\)
∴ \(x=-3sin\frac { 2\pi }{ T } t\quad or\quad x=-3sin\pi t\)
which is the equation of SHM
(2) Let B be any point on the circle of reference of fig. (b). From B, draw BN perpendicular on x-axis.
Then ∠BON = θ = ωt
∴ In ΔONB, cosθ = \(\frac{ON}{OB}\)
or ON = OB cos θ

∴ -x = 2 cos ωt
⇒ x=-2cos\(\frac{2\pi}{T}t\) =- 2 cos\(\frac{2\pi}{4}t\)
∴ x=- 2 cos \(\frac{\pi}{4}t\) which is equation of SHM

The given displacement function is
x(t) = A cos (ωt + \(\phi \)) ...(i)
At t 0, x(0) = 1 cm. Also, ω = \(\pi\)s^-1
∴ 1 = A cos(\(\pi\)\(\times\)0+\(\phi \))
⇒ A cos\(\phi \) = 1 ....(ii)
Also, differentiating eqn. (i) w.r.t. 't'.
v = \(\frac{d}{dt}x(t)\)=-Aωsin(ωt + \(\phi \)) ....(iii)
Now at t = 0, v = ω
∴ from eqn. (iii), ω = -Aωsin(\(\pi\)\(\times\)0 + \(\phi \))
or A sin \(\phi \)=-1 ....(iv)
Squaring and adding eqns. (ii) and (iv).
A² cos²\(\phi \)+ A² sin²\(\phi \) = 1² + 1² or A =\(\sqrt2\)cm
Dividing eqns. (ii) and (iv),
\(\frac { Asin\phi }{ Acos\phi } =\frac { -1 }{ 1 } \therefore tan\phi =-1\Rightarrow \phi =\frac { 3\pi }{ 4 } \)
If instead we use the sine function, i.e.,
x = B sin (ωt+ α),then v =\(\frac{d}{dt}\)Bωcos(ωt +  α)
∴ At t = 0, using x = 1and v = ω, we get 1 = B sin(ω\(\times\)0+α)
or B sin α = 1 ...(v)
and ω = Bωcos(ω\(\times\)0+α) or Bcosα = 1 ...(vi)
Dividing (v) by (vi),
tanα = 1 or α = \(\frac{\pi}{4}\) or \(\frac{5\pi}{4}\)
Squaring (v) and (vi), we get
B² sin²α + B² cos² α= 1² + 1²
⇒ B =​ \(\sqrt2\) cm

(a) (i) Given, function is
\(\sin { \omega t } -\cos { \omega t } \) = \(\sin { \omega t } -\sin { \left( \frac { \pi }{ 2 } -\omega t \right) } \) 
\(=2\sin { \left( \frac { \omega t+\frac { \pi }{ 2 } -\omega t }{ 2 } \right) } .\sin { \left( \frac { \omega t-\frac { \pi }{ 2 } +\omega t }{ 2 } \right) } \) 
Given function \(=2\sin { \left( \frac { \pi }{ 4 } \right) } .\sin { \left( \omega t-\frac { \pi }{ 4 } \right) } \)  
\(=\sqrt { 2 } \sin { \left( \omega t-\frac { \pi }{ 4 } \right) } \) 
This function represents a simple harmonic motion having period of \(T=\frac { 2\pi }{ \omega } \) and a phase angle \(\left( -\frac { \pi }{ 4 } \right) \) or \(\left( -\frac { 7\pi }{ 4 } \right) \).
(b) Periodic, but not SHM
The given function is:
sin3ωt = 143sin ωt - sin3 ωt
The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.(c) 3cos(\(\frac{\pi}{4}\)-2ωt) = 3 cos(2ωt -\(\frac{\pi}{4}\)) [∵ cos (- θ) = cos θ]
Clearly it represents SHM and its time period is 2π/2ω
(d) cos ωt + cos 3 ωt + cos 5 ωt. It represents the periodic but not S.H.M. Its time period is 2π/2ω
(e) \({ e }^{ -{ \omega }^{ 2 }{ t }^{ 2 } }\) It is an exponential function which never repeats itself. Therefore it represents non-periodic motion.
(f) 1 + ωt + ω²t²  also represents non periodic motion.