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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
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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 ______.
Acceleration of a particle executing S.H.M. at its mean position.
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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.
State the expression for the total energy of SHM in terms of acceleration.
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