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 simple harmonic motion with amplitude A and angular velocity ω. The ratio of maximum velocity to maximum acceleration is ______.
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.
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 ______.
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 ______.
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) ____________.
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 length of the second's pendulum is decreased by 0.3 cm when it is shifted from place A to place B. If the acceleration due to gravity at place A is 981 cm/s2, the acceleration due to gravity at place B is ______ (Take π2 = 10)
A body is executing S.H.M. Its potential energy is E1 and E2 at displacements x and y respectively. The potential energy at displacement (x + y) is ______.
A simple pendulum of length 'L' is suspended from a roof of a trolley. A trolley moves in horizontal direction with an acceleration 'a'. What would be the period of oscillation of a simple pendulum?
(g is acceleration due to gravity)
A block of mass 16 kg moving with velocity 4 m/s on a frictionless surface compresses an ideal spring and comes to rest. If force constant of the spring is 100 N/m then how much will be the spring compressed?
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?
A particle is performing SHM starting extreme position, graphical representation shows that between displacement and acceleration there is a phase difference of ______.
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 spring of force constant of 400 N/m is loaded with a mass of 0.25 kg. The amplitude of oscillations is 4 cm. When mass comes to the equilibrium position. Its velocity is ______.
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 expressions for the displacement, velocity and acceleration draw performing linear SHM, starting from the positive extreme position. Hence, their graphs with respect to time.
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 ______.
