There are three laws of transverse vibrations of stretched strings which are given as follows:
(i) The law of length : For a given wire with tension T (which is fixed) and mass 'per unit' length Il (fixed) the frequency varies inversely with the vibrating length. Therefore
\(f∝{1\over l}⇒f={C\over l}\)
⇒ l x f = C, where C is a constant
(ii) The law of tension: For a given vibrating length I (fixed) and mass per unit length Il (fixed) the frequency varies directly with the square root of the tension T,
f∝√T
⇒ f = A√T, where A is a constant
(iii) The law of mass : For a given vibrating length I (fixed) and tension T (fixed) the frequency varies inversely with the square root of the mass per unit length μ
\(f∝{1\over\sqrt\mu}\)
\(⇒f={B\over \sqrt{\mu}}\) where B is a constant
 

(i) Consider a massless spring with stiffness constant or force constant k attached to a ceiling.
(ii) Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring elongates by a length I.
(iii) Let F₁ Ibe the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system. When the system is under equilibrium,
F₁+mg = 0
(iv) But the spring elongates by small displacement l therefore,
F₁ ∝ 1=> F₁ = - kl
Substituting equation in equation, we get
-kl+ mg = 0
mg = kl
or
\(\frac { m }{ k } =\frac { l }{ g } \)
(v) Suppose we apply a very small external force is applied on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + 1 ) is
F₂∝(y+l)
F₂ = - k (y + l) == -ky - kl
Since, the mass moves up and down with \(\frac { { d }^{ 2 }y }{ dt^{ 2 } } \) acceleration by drawing the free body diagram for this case, we get \(-ky-kl+mg=m\frac { d^{ 2 }y }{ { dt }^{ 2 } } \)
(vi) The net force acting on the mass due to this stretching is
F = F₂ + mg
F = - ky - kl + mg
The gravitational force opposes the restoring force. Substituting equation in equation, we get
F = -ky - kl + kl = -ky
Applying Newton's law, we get
m\(\frac { { d }^{ 2 }y }{ dt^{ 2 } } \)=ky
\(\frac { { d }^{ 2 }y }{ dt^{ 2 } } =\frac { k }{ m } \)y
(v) The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as
  T=\(2\pi \sqrt { \frac { m }{ k } } \)
(vi) The time period can be rewritten using equation
\(2\pi \sqrt { \frac { m }{ k } } =2\pi \sqrt { \frac { l }{ g } } \)
The acceleration due to gravity g can be computed from the formula g = \(4{ \pi }^{ 2 }\left( \frac { l }{ { T }^{ 2 } } \right) \)ms^-1.

The first law of thermodynamics explains well the inter convertibility of heat and work. But it does not indicate the direction of change.
For example:
(a) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law, it is possible for the energy to flow from hot object to cold object or from cold object to hot object. But in nature the direction of heat flow is always from higher temperature to lower temperature.
(b) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted to the kinetic energy of the car. So the first law is not sufficient to explain many of natural phenomena. 

 Consider a capillary tube which is held vertically in a beaker containing water; the water rises in the capillary tube to a height h due to surface tension.
The surface tension force F_T, acts along the tangent at the point of contact .downwards and its reaction force upwards. Surface tension T, is resolved into two 'components
(i) Horizontal component T sinθ and
(ii) Vertical component T cosθ acting upwards, all along the whole circumference of the meniscus. Total upward force = (T cosθ) (2πr) = 2πrT cosθ where S is the angle of contact, r is the radius of the tube. Let p be the density of water and h be the height to which the liquid rises inside the tube. Then,  
(the volume of liquid column in the tube, V = (Volume of the liquid column of radius r height h)+
(Volume of liquid of radius r and height r - Volume of the hemisphere of radius r)

The upward force supports the weight of the liquid column above the free surface, therefore,
\(2\pi rT cos\theta=\pi r^{2} (h+\frac{1}{3}r)\rho g \Rightarrow T= \frac{r(h+\frac{1}{3}r)\rho g}{2 cos \theta}\)
If the capillary is a very fine tube of radius (i.e., radius is very small) then \(\frac{r}{3}\) can be neglected3
when it is compared to the height h. Therefore,
\(T=\frac{r\rho gh}{2 cos \theta}\)

Consider the Earth and mass system, with r, the distance between the mass m and the Earth's centre. Then the gravitational potential energy,
\(\mathrm{U}=-\frac{G M_{e} m}{r}\) ......(1)
Here r = R_e + h, where R_e is the radius of the Earth. h is the height above the Earth's surface.
\(\mathrm{U}_{\mathrm{c}}=-G \frac{M_{e} m}{\left(R_{e}+h\right)}\) .......(2)
If h << R_e, equation can be modified as
\(\mathrm{U} =-G \frac{M_{e} m}{\mathrm{R}_{e}\left(1+h / R_{e}\right)}  \)
\(\mathrm{U} =-G \frac{M_{e} m}{\mathrm{R}_{e}}\left(1+h / R_{e}\right)^{-1}\)  .......(3)
By using Binomial expansion and neglecting the higher order terms, we get
\(\mathrm{U}=-G \frac{M_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{R_{e}}\right)\)   ........(4)
We know that, for a mass m on the Earth's surface,
\(G \frac{M_{e} m}{\mathrm{R}_{e}}=m g \mathrm{R}_{e}\)    ........(5)
Substituting equation (5) in (4) we get
\(\mathrm{U}=-\mathrm{mg} \mathrm{R}_{e}+\mathrm{mgh}\)   .......(6)
It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂ then the potential energy at h₁ is
\(\mathrm{U}\left(\mathrm{h}_{1}\right)=-\mathrm{mg} \mathrm{} \mathrm{R}_{\mathrm{e}}+\mathrm{mgh}_{1}\)  .......(7)
and the potential energy at h₂ is
\(\mathrm{U}\left(\mathrm{h}_{2}\right)=-\mathrm{mg} \mathrm{} \mathrm{R}_{\mathrm{e}}+\mathrm{mgh}_{2}\)  .....(8)
The potential energy difference between h₁ and h₂ is
\(U\left(h_{2}\right)-U\left(h_{1}\right)=m g\left(h_{1}-h_{2}\right)\) .......(9)

(i) Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.
(ii) If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²
(iii) Let us consider a rigid body as shown in Figure. Its moment of inertia about an axis AB passing through the center of mass is I_C DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C For this, let us consider a point mass m on the body at position x from its center of mass.

(iv) The moment of inertia of the point mass about the axis DE is, m(x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
\(I=\sum { m\left( x+d \right) ^{ 2 } } \)
This equation could further be written as,
\(I=\sum { m\left( { x }^{ 2 }+{ d }^{ 2 }+2xd \right) } \)
\(I=\sum { \left( { mx }^{ 2 }+m{ d }^{ 2 }+2dmx \right) } \)
\(I=\sum { { mx }^{ 2 }+\sum { m{ d }^{ 2 } } +2d\sum { mx } } \)
(v) Here, \(\sum { mx^{ 2 } } \) is the moment of inertia of the body about the center of mass. Hence, 
I_C = \(\sum { mx= } 0\) because, x can take positive and negative values with respect to the axis AB. The summation \(\left( \sum { mx } \right) \) will be zero.
Thus, I = I_c + \(\sum { md^{ 2 } } \) = I_C + \(\left( \sum { m } \right) d^{ 2 }\)
(vi) Here, \(\sum { m } \) is the entire mass M of the object \(\left( \sum { m=M } \right) \)
I = I_C + Md²
Hence the parallel axis theorem is proved.

Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x-direction) on a frictionless horizontal.
      
 

Mass	Initial Velocity	Final Velocity
Mass m1	u1	v1
Mass m2	u2	v2

(i) In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂  For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

	Momentum of mass m1	Momentum of mass m2	Total linear momentum
Before collision	Pi1 = m1u1	Pi2 = m2u2	Pi = pi1 + Pi2  Pi = m1u1 + m2u2
After collision	Pf1 = m1v1	Pf2 = m2v2	Pf = Pf1 + Pf2  Pf = m1v1 + m2v2

From the law of conservation of linear momentum,
Total momentum before collision (p_i) = Totai momentum after collision (P_f)
Further,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v_s ...........(1)
or
m₁ (u₁ - v₁) = m₂ (v₂ - u₂) ...............(2)

	Kinetic energy of mass m1	Kinetic energy of mass m2	Total kinetic energy
			KEi = KEi1 + KEi2
Before collision	KEi1 =  \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }\)	KEi2 = \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 2 }^{ 2 }\)	KEi = \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }\) + \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 2 }^{ 2 }\) KEi = KEi1 + KEi2
After collision	KEf1 = \(\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 1 }^{ 2 }\)	KEf2 = \(\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 2 }^{ 2 }\)	KEf1 = \(\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 1 }^{ 2 }\) + \(\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 2 }^{ 2 }\)

For elastic collision,
Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f.
       \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }=\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ v }_{ 2 }^{ 2 }\)   ................(3)
After simplifying and rearranging the terms,
        \({ m }_{ 1 }\left( { u }_{ 1 }^{ 2 }-{ v }_{ 1 }^{ 2 } \right) ={ m }_{ 2 }\left( { v }_{ 2 }^{ 2 }-{ u }_{ 2 }^{ 2 } \right) \)
Using the formula a² - b² = (a + b) (a - b), we can rewrite the above equation as
        m₁ (u₁ + v₁) (u₁ - v₁) = m₂ (v₂ + u₂) (v₂ - u₂) .................(4)
Dividing equation (4) by (2) gives,
          \(\frac { { m }_{ 1 }\left( { u }_{ 1 }+{ v }_{ 1 } \right) \left( { u }_{ 1 }-{ v }_{ 1 } \right) }{ { m }_{ 1 }\left( { u }_{ 1 }-{ v }_{ 1 } \right) } =\frac { { m }_{ 2 }\left( { u }_{ 2 }+{ v }_{ 2 } \right) \left( { u }_{ 2 }-{ v }_{ 2 } \right) }{ { m }_{ 2 }\left( { u }_{ 2 }-{ v }_{ 2 } \right) } \)
   u₁ + v₁ = v₂ + u₂
    u₁ - u₂ = v₂ - v₁ .............(5)
Equation (5) can be rewritten as
 (u₁ - u₂) = -(v₁ - v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂+ u₂ - u₁ .........(6)
or
v₂ = u₁ + v₁ - u₂ ............(7)
To find the final velocities v₁ and v₂:
Substituting equation (7) in equation (2) gives the velocity of ml as
m₁ (u₁ - v_I) = m₂ (u₁ + v₁ - u₂ - u₂)
m₁ (u₁ - v₁) = m₂ (u₁ + v₁ - 2u₂)
m₁u₁ - m₁v₁ = m₂u₁ + m₂v₁ - 2m₂u₂
m₁u₁ - m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ - m₂)u₁ + 2m₂u₂ = (m₁ + m₂) v₁
or \({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) ............(8)
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of m₂ as
\({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) { u }_{ 2 }\) ............... (9)
Case 1:
When bodies has the same mass i.e., m₁ = m₂,
equation (8) \(\Rightarrow { v }_{ 1 }=\left( 0 \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { 2m }_{ 2 } } \right) { u }_{ 2 }\)
v₁ = u₂ ...........................(10)
equation (9)  \(\Rightarrow { v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { 2m }_{ 1 } } \right) { u }_{ 1 }+\left( 0 \right) { u }_{ 2 }\)
v₂ = u₁ ..........................(11)
The equations (10) and (11) show that in one dimensional elastic collision when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2:
When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u₂ = 0),
By substituting m₁ = m₂ = and u₂ = 0 in equations (8) and (9).
we get,
from equation (8) => v₁ = 0 (..................... 12)
from equation (9) => v₂ = u₁ ( .................. 13)
Equations (12) and (13) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3:
The first body is very much lighter than the second body
\(\left( { m }_{ 1 }<{ m }_{ 2 },\frac { { m }_{ 1 } }{ { m }_{ 2 } } <1 \right) \) then the ratio \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } = 0\) and also if the target is at rest (u₂ = 0)
Dividing numerator and denominator of equation (8) by m₂, we get
\({ v }_{ 1 }=\left( \frac { \frac { { m }_{ 1 } }{ { m }_{ 2 } } -1 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 2 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 1 }=\left( \frac { 0-1 }{ 0+1 } \right) { u }_{ 1 }\)
v₁ = - u₁ ............................(14)
Similarly,
Dividing numerator and denominator of equation (9) by m₂, we get 
\({ v }_{ 2 }=\left( \frac { 2\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 2 }=\left( 0 \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
v₂ = 0  ..............................(15)
The equation (14) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (15) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.
Case 4:
The second body is very much lighter than the first body
\(\left( { m }_{ 2 }<<{ m }_{ 1 },\frac { { m }_{ 2 } }{ { m }_{ 1 } } <<1 \right) \) then the ratio \(\frac { { m }_{ 2 } }{ { m }_{ 1 } } = 0\) and also if the target is at rest (u₂ = 0).
Dividing numerator and denominator of equation 8 by m₁ we get
\({ v }_{ 1 }=\left( \frac { 1-\frac { { m }_{ 2 } }{ { m }_{ 1 } } }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) { u }_{ 1 }+\left( \frac { 2\frac { { m }_{ 2 } }{ { m }_{ 1 } } }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) \left( 0 \right) \)
\({ v }_{ 1 }=\left( \frac { 0-1 }{ 0+1 } \right) { u }_{ 1 }+\left( \frac { 0 }{ 1+0 } \right) \left( 0 \right) \)
v₁  = u₁  .....................(16)
Dividing numerator and denominator of equation (14) by m₁ we get
\({ v }_{ 2 }=\left( \frac { 2 }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) { u }_{ 1 }+\left( \frac { \frac { { m }_{ 2 } }{ { m }_{ 1 } } -1 }{ 1+\frac { { m }_{ 2 } }{ { m }_{ 1 } } } \right) \left( 0 \right) \)
\({ v }_{ 2 }=\left( \frac { 2 }{ 1+0 } \right) { u }_{ 1 }\)
v₂ = 2u₁   ....................(17)
The equation (16) implies that the first body which is heavier continues to move with the same initial velocity. The equation (17) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.

Consider an inclined plane on which an object is placed as shown in the figure, Let the angle which this plane makes with the horizontal be ፀ. For small angle of ፀ, the object may not slide down. As ፀ is increased, for a particular value of ፀ, the object begins to slide down. This value is called angle of repose. Hence, the angle of repose is the angle of the inclined plane with the horizontal such that an object placed on it begins to slide.

Consider a uniform circular motion of a point object moving on a circular path of radius r. In uniform circular motion, the velocity vector turns continuously without changing its magnitude (speed). Let v be its velocity.
The velocity is tangential at every point in the circle. But the acceleration 'a' is acting towards the centre of the circle. This is called centripetal acceleration. It always points towards the centre of the circle.
Let the direction of position and velocity vectors shift through the same angle θ in a small interval of time \(\triangle t\) as shown in figure. For uniform circular motion \(r=|\overrightarrow{r_1}|={r_2}| \) and \(vr=|\overrightarrow{v_1}|=\overrightarrow{v_2}|\)

If the particle moves from position vector \(\overrightarrow{r_1}\) to \(\overrightarrow{r_2}\) , the displacement is \(\triangle{\overrightarrow{r}}={\overrightarrow{r}}_{2}-{\overrightarrow{r}}_{1}\) and change in velocity from \(\overrightarrow{v_1}\) to \(\overrightarrow{v_2}\) is \(\triangle={\overrightarrow{v}}_{1}={\overrightarrow{v}}_{2}-{\overrightarrow{v}}_{1}\) then the magnitude of the displacement \(\triangle r \ and \ of \triangle v\) satisfy the following relation:
\({{\triangle r}\over{r}}=-{{\triangle v}\over{v}}=\theta\)
Here the negative sign implies that \(\triangle r\) points radially inward, towards the centre of the circle
\(\triangle v=-v.{{\triangle r}\over{r}}\)
Then \(a={{\triangle r}\over{\triangle t}}={{r}\over{r}}\left( {{\triangle r}\over{\triangle t}} \right)=-{{v}^{2}{}\over{r}}\)
For uniform circular motion v = \(\omega\)r. where \(\omega\) - angular velocity of the object about the centre. Then the centripetal acceleration can be written as
\({a=-\omega^2 r}\)

Dimension formula for 
 \(\frac { 1 }{ 2 } \)mv² = [M][LT^-1]² = [ML²T^-2]
Dimension formula for 
mgh = [M][LT^-2][L] = [ML²T^-2]
[ML²T^-2] = [ML²T^-2]
Both sides are dimensionally the same, hence the equations \(\frac { 1 }{ 2 } \)mv² = mgh is dimensionally correct