Advertisements
Advertisements
प्रश्न
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.]
Advertisements
उत्तर १
The displacement equation for an oscillating mass is given by:
x = `A cos (omega t + theta)`
Where,
A is the amplitude
x is the displacement
θ is the phase constant
Velocity, `v = (dx)/(dt) = -Aomega sin(omega t + theta)`
At t = 0, x = x0
x0 = Acosθ = x0 … (i)
And, `(dx)/(dt) = -v_0 = Aomegasin theta`
`A sin theta = v_0/omega` ...(ii)
Squaring and adding equations (i) and (ii), we get:
`A^2(cos^2 theta + sin^2 theta)= x_0^2 + ((v_0^2)/(omega^2))`
`:. A= sqrt(x_0^2 + (v_0/omega)^2)`
Hence, the amplitude of the resulting oscillation is `sqrt(x_0^2 + (v_0/omega)^2)`
उत्तर २
x = `alpha cos (omegat + theta)`
`v = (dx)/(dt) = -aomega sin(omegat = theta)`
When t = 0, x = `x_0` and `dx/dt = -v_0`
x0 = `a cos theta`
and `-v_0 = -a omega sin theta` or a `sin theta = v_0/omega`
Squaring and adding (i) and (ii) we get
`a^2(cos^2 theta + sin^2 theta) = x_0^2 + (v_0^2)/(omega^2)`
or `a = sqrt(x_0^2 + v_0^2/omega^2)`
APPEARS IN
संबंधित प्रश्न
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.
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`
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?
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:
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?
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?
Define practical simple pendulum
Show that, under certain conditions, simple pendulum performs the linear simple harmonic motion.
If the particle starts its motion from mean position, the phase difference between displacement and acceleration is ______.
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 ______.
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
The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?
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 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.
