Advertisements
Advertisements
Question
The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is ______.
Options
`1/(2 pi) sqrt((l cos theta)/g)`
`1/(2 pi)sqrt((l sin theta)/g)`
`4 pi sqrt((l cos theta)/(4 g))`
`4 pi sqrt((l tan theta)/g)`
Advertisements
Solution
The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is `bbunderline(4 pi sqrt((l cos theta)/(4 g)))`.
Explanation:
For a conical pendulum, the vertical component of tension balances weight:
T cos θ = mg
= `4 pi sqrt ((l cos theta)/g)`
= `4 pi (1/2) sqrt((l cos theta)/g)`
= `2 pi sqrt((l cos theta)/g)`
Using centripetal force and angular motion relations, we get the time period:
T = `2 pi sqrt((l cos theta)/g)`
∴ The expression `4 pi sqrt ((l cos theta)/g)` is mathematically identical to the standard period formula for a conical pendulum.
RELATED QUESTIONS
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 phase difference between displacement and acceleration of a particle performing S.H.M. is _______.
(A) `pi/2rad`
(B) π rad
(C) 2π rad
(D)`(3pi)/2rad`
A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer the following questions:
A time period of a particle in SHM depends on the force constant k and mass m of the particle: `T = 2pi sqrt(m/k)` A simple pendulum executes SHM approximately. Why then is the time
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.
Answer the following questions:
A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
Answer the following questions:
What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
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).
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x = acos (ωt+θ) and note that the initial velocity is negative.]
If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be ______.
The period of oscillation of a simple pendulum of constant length at the surface of the earth is T. Its time period inside mine will be ______.
A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s2. The period of oscillation is ______.
Which of the following statements is/are true for a simple harmonic oscillator?
- Force acting is directly proportional to displacement from the mean position and opposite to it.
- Motion is periodic.
- Acceleration of the oscillator is constant.
- The velocity is periodic.
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 tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.
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:
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is `x/2` times its original time period. Then the value of x is ______.
