Mathematics Definition & Proof Based Problems

Definition & Proof Based Problems

Definition & Proof Based Problems
Q 3136878772

What is simple pendulum ? Show that the motion of the pendulum is S.H.M. and hence deduce an expression for the time period of pendulum. Also define second's pendulum.


In simple pendulum a heavy point mass body i.e., bob suspended by a weightless inextensible and perfectly flexible string from a rigid support about which it is free to oscillate.

m = mass of bob
l = length of pendulum

When bob is displaced to point P through
small angle `theta`.
Various forces acting on the bob is
(i) weight mg of the bob acting vertically
(ii) Tension T in string along PS.
Resolving mg into two components
(a) mg cos `theta` opposite to tension T .
(b) mg sin `theta` directed towards `theta`.If the string remains taut
`T= mg cos theta`
The force mg sin `theta` tends to bring back the bob to its mean position `theta`. . . Restoring force acting on bob is
`F=- mg sin theta` - ve sign shows force is directed towards mean position. If `theta` is small, then sin 0 = 0 = (arc OP)/l = x/l
`F = - mg theta` = - mg .. . (i)
`F prop` displacement `(x)` and F is directed
towards mean position `theta`.
In S.H.M, Restoring force

`F =- kx`

Comparing (i) and Ui)

`k = (mg)/x`

inertia factor `=` mass of bob `= m`

`T = 2pi (sqrttext(inertia factor)/text(spring factor)`

`= 2pi sqrt(m/(mgll)) = 2pi qsrt(l/g)`

Simple pendulum whose time period of vibrations is two seconds is called a second's pendulum.
Q 3106867778

Prove that if a liquid taken in a U-tube is disturbed from the state of equilibrium, it will oscillate simple harmonically. Find expressions for the angular frequency and time period.


The restoring force,
`F =` weight of liquid column of the height `2y`

`=> F = - ("volume") xx "density" xx g`
`= - (A xx`
`=> F = - 2Apg.y ... (i)`
where `A =` Area of cross-section of the tube

`p =` density of mercury

Time period

`T = 2pi sqrt(m/k)`
`= 2pi sqrt(m/(2Apg))`

Let `L =` length of the whole mercury column therefore, mass of mercury

`m = "volume" xx "density" = A.L.p.`

`T = 2pi sqrt(m/k) = 2pi sqrt(A.L.p)/(2Apg) = 2pi sqrt(L/(2g)` where L

is the total length of mercury column of `L = 2h.`

Where h is the height of mercury column in U-tube. It shows that mercury column executes S.H.M.

Frequency, `v = 1/T = 1/(2pi) sqrt(2g)/(L)`

So, angular frequency

`omega = 2piv = 2pi xx 1/(2pi) xx 1/(2pi) sqrt(2g)/(L) = sqrt(2g)/(L)`
Q 3166356275

State force law for a simple harmonic motion.


Force `F prop - x => F = kx => F = - momega^2x`
Q 3116778670

What is Simple Harmonic Motion? Show that in S.H.M., acceleration is directly proportional to its displacement at a given instant.


Simple Harmonic Motion :
(i) Motion is always directed towards a fixed point or equilibrium point.
(ii) Motion being represented by bounded trigonometric functions.
(iii) Acceleration is directly proportional to negative of displacement, i.e., a `prop-· x` Equation for S.H.M.

Acceleration `= - omega^2 x`

`=> (d^2x)/(dt^2) + omega^2x = 0 omega = 2pif`

w is angular frequency (radian/sec), f is linear frequency `(s-1)` or (hertz)
Q 1980112017

Show that the motion of a particle represented by

`y = sin omegat - cos omega t`

is simple harmonic with a period of `2pi /omega`

NCERT Exemplar

We have to convert the given combination of two harmonic functions to a single harmonic
(sine or cosine) function.
Given, displacement function

`y = sinomegat - cos omegat`

`= sqrt2(1/sqrt2.sinomegat-1/sqrt2.cosomegat)`

`= sqrt2[cos(pi/4).sinomegat - sin(pi/4).cos omegat]`

`= sqrt2[sin(omega-pi/4)] = sqrt2[sin(omegat-pi/4)]`

Comparing with standard equation

`y = asin(omegat+phi),` we get `= omega = (2pi)/(T) => T = (2pi)/(omega)`

Clearly, the function represents SHM with a period `T = (2pi)/omega`