Advertisements
Advertisements
प्रश्न
Answer in brief.
Using differential equations of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.
State the differential equation of linear S.H.M. Hence, obtain the expression for:
- acceleration
- velocity
Advertisements
उत्तर
Differential equation of SHM
∴ `(d^2x)/(dt^2) + omega^2.x = 0` ............(i)
(a) Obtaining expression for acceleration:
`(d^2x)/(dt^2) = -omega^2.x` ........[From (i)]
But `a = (d^2x)/(dt^2)` is the acceleration of the particle performing SHM.
∴ a = `-omega^2.x`
This is the expression for acceleration.
(b) Obtaining expression for velocity:
`(d^2x)/(dt^2) = -omega^2.x` ..............[From (i)]
∴ `d/dt((dx)/(dt)) = -omega^2.x`
∴ `(d"v")/(dt) = -omega^2.x`
∴ `(d"v")/(dx)(dx)/(dt) = -omega^2.x`
∴ v`(d"v")/(dx) = -omega^2.x`
∴ v.dv = -ω2.x.dx
Integrating both sides, we get
`int"v".d"v" = -omega^2intx.dx`
∴ `"v"^2/2 = -omega^2.x^2/2 + C` ...........(iii)
Where C is the constant of integration.
Let A be the maximum displacement (amplitude) of the particle in SHM.
When the particle is at an extreme position, velocity (v) is zero,
Thus, at x = ±A, v = 0,
Substituting v = 0 and x = ±A in equation (ii), we get
∴ `0 = -omega^2. A^2/2 + C`
∴ C = `omega^2.A^2/2`
Using the value of C in equation (ii), we get
∴ `"v"^2/2 = -omega^2.x^2/2 + omega^2.A^2/2`
∴ `"v"^2 = omega^2(A^2 - x^2)`
∴ v = `±omegasqrt(A^2 - x^2)`
This is the expression for velocity.
संबंधित प्रश्न
The light of wavelength '`lambda`'. incident on the surface of metal having work function `phi` emits the electrons. The maximum velocity of electrons emitted is ______.
[c = velocity of light, h = Planck's constant, m = mass of electron]
For a particle performing SHM when displacement is x, the potential energy and restoring force acting on it is denoted by E and F, respectively. The relation between x, E and F is ____________.
The displacement of a particle from its mean position (in metre) is given by, y = 0.2 sin(10 πt + 1.5π) cos(10 πt + 1.5π).
The motion of particle is ____________.
In U.C.M., when time interval δt → 0, the angle between change in velocity (δv) and linear velocity (v) will be ______.
A wheel of M.I. 50 kg m2 starts rotating on applying a constant torque of 200 Nm. Its angular velocity after 2.5 second from the start is ______.
A particle is moving along a circular path of radius 6 m with a uniform speed of 8 m/s. The average acceleration when the particle completes one-half of the revolution is ______.
The relation between time and displacement for two particles is given by Y1 = 0.06 sin 27`pi` (0.04t + `phi_1`), y2 = 0.03sin 27`pi`(0.04t + `phi_2`). The ratio of the intensity of the waves produced by the vibrations of the two particles will be ______.
The phase difference between the instantaneous velocity and acceleration of a particle executing S.H.M is ____________.
If 'α' and 'β' are the maximum velocity and maximum acceleration respectively, of a particle performing linear simple harmonic motion, then the path length of the particle is _______.
The distance covered by a particle undergoing SHM in one time period is (amplitude = A) ____________.
A particle executing S.H.M. has amplitude 0.01 m and frequency 60 Hz. The maximum acceleration of the particle is ____________.
The maximum speed of a particle in S.H.M. is 'V'. The average speed is ______
The bob of a simple pendulum is released at time t = 0 from a position of small angular displacement. Its linear displacement is ______.
(l = length of simple pendulum and g = acceleration due to gravity, A = amplitude of S.H.M.)
The displacement of a particle in S.H.M. is x = A cos `(omegat+pi/6).` Its speed will be maximum at time ______.
The displacements of two particles executing simple harmonic motion are represented as y1 = 2 sin (10t + θ) and y2 = 3 cos 10t. The phase difference between the velocities of these waves is ______.
The displacement of a particle is represented by the equation `y = 3 cos (pi/4 - 2ωt)`. The motion of the particle is ______.
In figure, a particle is placed at the highest point A of a smooth sphere of radius r. It is given slight push and it leaves the sphere at B, at a depth h vertically below A, such that h is equal to ______.
The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin (0.2 t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to `1/3` of its amplitude from its mean position is ______.
A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m–1/2s–1. The change in kinetic energy during its displacement from x = 0 to x = 2 m is ______.
In the given figure, a = 15 m/s2 represents the total acceleration of a particle moving in the clockwise direction on a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is ______.
For a particle performing circular motion, when is its angular acceleration directed opposite to its angular velocity?
State the expression for the total energy of SHM in terms of acceleration.
A particle executing SHM has velocities v1 and v2 when it is at distance x1 and x2 from the centre of the path. Show that the time period is given by `T=2pisqrt((x_2^2-x_1^2)/(v_1^2-v_2^2))`
A body is suspended from the two light springs separately, the periods of vertical oscillations are \[T_1\] and \[T_2\]. The same body is suspended from two springs connected in series first and then in parallel. The periods of oscillations in both cases respectively are ______.
A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance `(2 A)/3` from equilibrium position. The new amplitude of the motion is ______.
