Let the equation of oscillation be given by
Velocity would be given as
Kinetic energy at an instant is given by
Time Period is given by
The Average Kinetic Energy would be given as follows
The potential energy at an instant T is given by
The Average Potential Energy would be given by
We can see Kav = Uav

At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.
Let the amplitude be A
The angular frequency of a spring-mass system is always equal to
Therefore

A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 0 cm

A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 3 cm

A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 5 cm

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is
The period of Torsional oscillations would be

The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.
For damping factor b we have
x=x0/2
t=0.77s
m=750 kg

Mass of automobile (m) = 3000 kg
There are a total of four springs.
Compression in each spring, x = 15 cm = 0.15 m
Let the spring constant of each spring be k

Let the initial volume and pressure of the chamber be V and P.
Let the ball be pressed by a distance x.
This will change the volume by an amount ax.
Let the change in pressure be
Let the Bulk's modulus of air be K.
This pressure variation would try to restore the position of the ball.
Since force is restoring in nature displacement and acceleration due to the force would be in different...

Let the height of each mercury column be h.
The total length of mercury in both the columns = 2h.
Let the cross-sectional area of the mercury column be A.
Let the density of mercury be
When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.
Weight of this difference is
This weight drives the rest of the entire column to the original...

Let the cork be displaced by a small distance x in downwards direction from its equilibrium position where it is floating.
The extra volume of fluid displaced by the cork is Ax
Taking the downwards direction as positive we have
Comparing with a=-kx we have

Acceleration due to gravity = g (in downwards direction)
Centripetal acceleration due to the circular movement of the car = ac
(in the horizontal direction)
Effective acceleration is
The time period is T'

While free falling the effective value of g inside the cabin will be zero and therefore the frequency of oscillation of a simple pendulum would be zero i.e. it would not vibrate at all because of the absence of a restoring force.

The watch must be using an electrical circuit or a spring system to tell the time and therefore free falling would not affect the time his watch predicts.

In reaching the result we have assumed sin(x/l)=x/l. This assumption is only true for very small values of x . Therefore it is obvious that once x takes larger values we will have deviations from the above-mentioned value.

In case of spring, the spring constant is independent of the mass attached whereas in case of a pendulum k is proportional to m making k/m constant and thus the time period comes out to be independent of the mass of the body attached.

The time period of a simple pendulum of length l executing S.H.M is given by
ge = 9.8 m s-2
gm = 1.7 m s-2
The time period of the pendulum on the surface of Earth is Te = 3.5 s
The time period of the pendulum on the surface of the moon is Tm

Amplitude of SHM = 0.5 m
angular frequency is
If the equation of SHM is given by
The velocity would be given by
The maximum speed is therefore

(b).(a) In Fig, (a) we have
F=-kx
ma=-kx
(b) In fig (b) the two equal masses will be executing SHM about their centre of mass. The time period of the system would be equal to a single object of same mass m attached to a spring of half the length of the given spring (or undergoing half the extension of the given spring while applied with the same force)
Spring constant of such a spring would...

(a) Let us assume the maximum extension produced in the spring is x.
At maximum extension
(b) Let us assume the maximum extension produced in the spring is x. That is x/2 due to force towards left and x/2 due to force towards right

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