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.

Solution:

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
downward.
(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.

Solution:

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

=> F = - ("volume") xx "density" xx g
= - (A xx -2y.pg)
=> 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.

Solution:

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.

Solution:

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
Solution:

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