Advertisements
Advertisements
प्रश्न
The cylindrical piece of the cork of density of base area A and height h floats in a liquid of density `rho_1`. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
`T = 2pi sqrt((hrho)/(rho_1g)`
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Advertisements
उत्तर १
Base area of the cork = A
Height of the cork = h
Density of the liquid = `rho_1`
Density of the cork = ρ
In equilibrium:
Weight of the cork = Weight of the liquid displaced by the floating cork
Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
Up-thrust = Restoring force, F = Weight of the extra water displaced
F = –(Volume × Density × g)
Volume = Area × Distance through which the cork is depressed
Volume = Ax
∴ F = – A x `rho_1` g … (i)
According to the force law:
F = kx
`k = F/x`
Where, k is a constant
`k = F/x = -Arho_1g` ...(ii)
The time period of the oscillations of the cork:
`T = 2pi sqrt(m/k)` ...(iii)
Where
m = Mass of the cork
= Volume of the cork × Density
= Base area of the cork × Height of the cork × Density of the cork
= Ahρ
Hence, the expression for the time period becomes:
`T = 2pi sqrt((Ahrho)/(Arho_1g)) = 2 pi sqrt((hrho)/(rho_1g)`
उत्तर २
Say, initially in equilibrium, y height of cylinder is inside the liquid. Then,
Weight of the cylinder = upthrust due to liquid displaced
`:. Ahrhog = Ayrho_1g`
When the cork cylinder is depressed slightly by `triangle y` and released, a restoring force, equal to additional upthrust, act on it. The restoring force is
`F = A(y + triangle y) rho_1 g - Ayrho_1g = Arho_1g triangle y`
:. Acceleration, `a = F/m = (Arho_1g triangle y)/(Ahrho) = (rho_1g)/(hrho). triangle y ` and the acceleration is directed in a direction opposite to `triangle y`. Obviously, as `a prop - triangle y`, the motion of cork cylinder is SHM, whose time period is given by
`T = 2pi sqrt("displacement"/"accelertion")`
`=2pi sqrt((triangle y)/a)`
`= 2pi sqrt((hrho)/(rho_1g))`

APPEARS IN
संबंधित प्रश्न
The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is ______.
If the metal bob of a simple pendulum is replaced by a wooden bob of the same size, then its time period will.....................
- increase
- remain same
- decrease
- first increase and then decrease.
When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length of the pendulum.
The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)
Answer the following questions:
The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than `2pisqrt(1/g)` Think of a qualitative argument to appreciate this result.
Show that motion of bob of the pendulum with small amplitude is linear S.H.M. Hence obtain an expression for its period. What are the factors on which its period depends?
Show that, under certain conditions, simple pendulum performs the linear simple harmonic motion.
A simple pendulum has a time period of T1 when on the earth's surface and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of `"T"_2 // "T"_1` is ______.
The relation between acceleration and displacement of four particles are given below: Which one of the particles is executing simple harmonic motion?
Two identical springs of spring constant K are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force

Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.

A body of mass m is situated in a potential field U(x) = U0 (1 – cos αx) when U0 and α are constants. Find the time period of small oscillations.
Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical. What is the phase difference between the pendulums?
A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. `T = 2πsqrt(m/(Apg))` where m is mass of the body and ρ is density of the liquid.
A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Figure). The amplitude is θ0. The string snaps at θ = θ0/2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θo to be small so that sin θo = θo and cos θo = 1.

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:

A pendulum of mass m and length ℓ is suspended from the ceiling of a trolley which has a constant acceleration a in the horizontal direction as shown in the figure. Work done by the tension is ______.
(In the frame of the trolley)

