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
A particle is performing S.H.M. of amplitude 5 cm and period of 2s. Find the speed of the particle at a point where its acceleration is half of its maximum value.
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 body performing a simple harmonic motion has potential energy 'P1' at displacement 'x1' Its potential energy is 'P2' at displacement 'x2'. The potential energy 'P' at displacement (x1 + x2) is ________.
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 body of mass 5 g is in S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/s. Its velocity will be 50 cm/s at a distance of, ____________.
The displacement of a particle is 'y' = 2 sin `[(pit)/2 + phi]`, where 'y' is cm and 't' in second. What is the maximum acceleration of the particle executing simple harmonic motion?
(Φ = phase difference)
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 the particle performing S.H.M. is given by x = 4 sin πt, where x is in cm and t is in second. The time taken by the particle in second to move from the equilibrium position to the position of half the maximum displacement, is ______.
`[sin30^circ=cos60^circ=0.5, cos30^circ=sin60^circ=sqrt3/2]`
The displacement of a particle in S.H.M. is x = A cos `(omegat+pi/6).` Its speed will be maximum at time ______.
A body perform linear simple harmonic motion of amplitude 'A'. At what displacement from the mean position, the potential energy of the body is one fourth of its total energy?
The displacement of a particle is represented by the equation `y = 3 cos (pi/4 - 2ωt)`. The motion of the particle is ______.
A particle performs linear SHM at a particular instant, velocity of the particle is 'u' and acceleration is a while at another instant velocity is 'v' and acceleration is 'β (0 < α < β). The distance between the two position 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 ______.
A particle of mass 5 kg moves in a circle of radius 20 cm. Its linear speed at a time t is given by v = 4t, t is in the second and v is in ms-1. Find the net force acting on the particle at t = 0.5 s.
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 ______.
Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = `5sin((pit)/3)`m.
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 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 ______.
