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
\(T=2\pi\sqrt{y\over |a|}=2\pi\sqrt{(k_1+k_2)M\over k_1k_2}\)
\(T=2\pi\sqrt{\left({1\over k_1}+{1\over k_2}\right)M}\)

For definition, see text. Let x₁ be the extension produced in the spring when it is loaded with mass m₂ alone and x₂ be the further extension when mass ml is added to mass m₂ so that x = x₁ + x₂ is the total extension produced by m₁ + m₂: (see Fig). Thus we have, For equilibrium state of m₂•

m₂ g = k x₁     ...(1)
For equilibrium state of (m₁ + m₂)
(m₂ + m₂) g = k(x) = k(x₁ + x₂)    ...(2)
When the mass m₁ is removed, the mass m₂ will move upwards under the unbalanced force = m₁g. Hence
Restoring force (F) on
m₂ = -m₁ g
Substracting (1) and (2) we have
m₁g = k x₂  ...(3)
Hence, Restoring force on
m₂ = -  k x₂
∴ acceleration of \(m_2={F\over m_2}=-{k\over m^2}x_2\)
∴ accelaration ∝: - displacement
Angular frequency is \(ω=\sqrt{k\over m_2}\)
∴ Frequency of oscillation is \(n={ω\over 2\pi}={1\over 2\pi}\sqrt{k\over m_2}\)
It is clear that A is the equilibrium postion of m₂ and Bits maximum displacement position. Hence AB = x₂ is the amplitude of oscillation of m₂ which from Eq. (3) is given by
Amplitude =  \(x_2={m_1g\over k}\)

Let the particle executingSHMstarts oscillating from its mean position. Then displacement equation is
x = A sin ωt
∴ Particle velocity, v = Aພcosωt
∴ Instantaneous K.E., K = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } { mA }^{ 2 }{ \omega }^{ 2 }{ cos }^{ 2 }\omega t\)
∴ Average value of K.E. over one complete cycle
\({ K }_{ av }=\frac { 1 }{ T } \int _{ o }^{ T }{ \frac { 1 }{ 2 } { mA }^{ 2 }{ \omega }^{ 2 }{ cos }^{ 2 }\omega tdt=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 2T } = } \int _{ 0 }^{ T }{ { { cos }^{ 2 }\omega tdt } } \)
\(=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 2T } \int _{ 0 }^{ T }{ \frac { (1+cos2\omega t) }{ 2 } dt } \)
\(=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 4T } { \left[ t+\frac { sin2\omega t }{ 2\omega } \right] }_{ 0 }^{ T }\)
\(=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 4T } \left[ (T-0)+\left( \frac { sin2\omega t-sin0 }{ 2\omega } \right) \right] \)
\(=\frac { 1 }{ 4 } { mA }^{ 2 }{ \omega }^{ 2 }...(i)\)
Again instantaneous P.E., U = \(\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } { m\omega }^{ 2 }{ x }^{ 2 }=\frac { 1 }{ 2 } { m\omega }^{ 2 }{ A }^{ 2 }{ sin }^{ 2 }\omega t\)
∴ Average value of P.E. over one complete cycle
\(U_{ av }=\frac { 1 }{ T } \int _{ o }^{ T }{ \frac { 1 }{ 2 } { m\omega }^{ 2 }{ A }^{ 2 }{ sin }^{ 2 }\omega t=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 2T } = } \int _{ 0 }^{ T }{ { { sin }^{ 2 }\omega tdt } } \)
\(=\frac { { m\omega }^{ 2 }{ A }^{ 2 } }{ 2T } \int _{ 0 }^{ T }{ \frac { (1-cos2\omega t) }{ 2 } dt } \)
\(=\frac { { m\omega }^{ 2 }{ A }^{ 2 } }{ 4T } { \left[ t-\frac { sin2\omega t }{ 2\omega } \right] }_{ 0 }^{ T }\)
\(=\frac { { m\omega }^{ 2 }{ A }^{ 2 } }{ 4T } \left[ (T-0)-\left( \frac { sin2\omega t-sin0 }{ 2\omega } \right) \right] \)
\(=\frac { 1 }{ 4 } { m\omega }^{ 2 }A^{ 2 }...(ii)\)
Simple comparison of (i) and (ii), shows that
\(K_{ av }=U_{ av }=\frac { 1 }{ 4 } { m\omega }^{ 2 }A^{ 2 }\) 

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 = \(\frac { 1 }{ T } =\frac { A }{ 2\pi } \sqrt { \frac { E }{ mV } } \) 

 Say, initially in equilibrium, y height of cylinder is inside the liquid. Then, Weight of the cylinder = upthrust due to liquid displaced
∴ Ahρg = Ayρ_lg
When the cork cylinder is depressed slightly by Δy and released, a restoring force, equal to additional upthrust, acts on it. The restoring force is
F = A(y +Δy) ρ_lg - Ayρ_lg = Aρ_lgΔy
∴ Acceleration \(a=\frac { F }{ m } =\frac { A{ \rho }_{ 1 }g\Delta y }{ Ah\rho } =\frac { { \rho }_{ 1 }g }{ h\rho } .\Delta y\)and the acceleration is directed in a direction opposite to Δy. Obviously, as a∞-Δy, the motion of cork cylinder is SHM, whose time period is given by

\(T=2\pi \sqrt { \frac { displacement }{ acceleration } } \)
\(=2\pi \sqrt { \frac { \Delta y }{ a } } \)
\(=2\pi \sqrt { \frac { h\rho }{ { \rho }_{ 1 }g } } \) 

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

Given that,
Spring constant k =1200 Nm^-1
mass m=3 kg
The distance at which mass is pulled from its equilibrium x=0.2 cm
(i) Frequency,
\( f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \)
\( =\frac{1}{2 \times 3.142} \times \sqrt{\frac{1200}{3}} \)
\( f=3.18 \mathrm{~Hz}\)
(ii) The acceleration is given by
\(a=-\omega^2 x=-\frac{k}{m} x\)
or \(\left|a_{\max }\right|=\frac{k}{m}\left|x_{\max }\right|\)
i.e., acceleration will be maximum, when x is maximum.
i.e., \(\mathrm{x}_{\max }=\mathrm{A}=0.02 \mathrm{~m}\)
\(\therefore \mathrm{a}=\frac{1200}{3} \times 0.02=8.0 \mathrm{~ms}^{-2}\)
(ii) The maximum speed of the mass is given by \( v=A \omega=A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}} =0.40 \mathrm{~ms}^{-1} \)

\(V^2=\omega^2\left(r^2-y^2\right)\)
(i) \(6^2=\omega^2\left(r^2-4^2\right)\)
(ii) \(\omega^2\left(r^2-3^2\right)\)
or \(\frac{64}{36}=\frac{r^2-9}{r^2-16}\)
On solving, \(r= \pm 5 m\) and \(\omega=2 s^{-1}\)
For the given displacement, \(x=\frac{r}{2}, t=\) ?
Asx \(x=r \cos \omega t\)
\(\therefore \frac{r}{2}=r \cos 2 t \text { or } \cos 2 t=\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)\)
or \(2 t=\frac{\pi}{3}\) or \(t=\frac{\pi}{6} s\)

Formula used: \(A_R=A \sqrt{2(1+\cos \delta}\)
Given: \(A_1=A_2=A_R=A\)
The resultant amplitude is also A.
If the phase difference between two motions is \(\delta\)
\( A=\sqrt{A^2+A^2+2 A \cdot A \cos \delta}\)
\( =A \sqrt{2(1+\cos \delta} \)
\( \Rightarrow A=2 A \cos \frac{\delta}{2} \)
\( \cos \frac{\delta}{2}=\frac{1}{2} \)
\( \Rightarrow \delta=\frac{2 \pi}{3} \)