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.

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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)`

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Solution:

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

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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)

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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`

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