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प्रश्न
Using the differential equation of linear S.H.M., obtain an expression for acceleration, velocity, and displacement of simple harmonic motion.
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उत्तर
- Expression for acceleration in linear S.H.M:
a. From the differential equation,
`("d"^2"x")/"dt"^2 + ω^2"x" = 0`
`("d"^2"x")/"dt"^2 = -ω^2"x"` ….(1)
b. But, linear acceleration is given by,
`("d"^2"x")/"dt"^2 = "a"` ...….(2)
From equations (1) and (2),
a = −ω2x .........….(3)
Equation (3) gives acceleration in linear S.H.M. - Expression for velocity in linear S.H.M:
a. From the differential equation of linear S.H.M
`("d"^2"x")/"dt"^2 = -ω^2"x"`
∴ `"d"/"dt"("dx"/"dt") = -ω^2"x"`
∴ `"dv"/"dt" = -ω^2"x"` ......`(∵ "dx"/"dt" = "v")`
∴ `"dv"/"dx"."dx"/"dt" = -ω^2"x"`
∴ `"v""dv"/"dx" = -ω^2"x"` ......`(∵ "dx"/"dt" = "v")`
∴ v dv = −ω2x dx ...........(4)
b. Integrating both sides of equation (4),
∫v dv = ∫-ω2x dx
`"v"^2/2 = -(ω^2"x"^2)/2` + C ............(5)
where, C is the constant of integration.
c. At extreme position, x = ± A and v = 0.
Substituting these values in equation (5),
0 = `-(ω^2"A"^2)/2 + "C"`
∴ C = `(ω^2"A"^2)/2` .............(6)
d. Substituting equation (6) in equation (5),
`"v"^2/2 = -(ω^2"x"^2)/2 + (ω^2"A"^2)/2`
∴ v2 = ω2A2 −ω2x2
∴ v2 = ω2(A2 - x2)
∴ v = ± ω`sqrt("A"^2 - "x"^2)`
This is the required expression for velocity in linear S.H.M. - Expression for displacement in linear S.H.M:
a. From the differential equation of linear S.H.M, velocity is given by,
v = ω`sqrt("A"^2 - "x"^2)` .....(1)
But, in linear motion, v = `"dx"/"dt"` ......(2)
From equation (1) and (2),
`"dx"/"dt" = ωsqrt("A"^2 - "x"^2)`
∴ `"dx"/sqrt("A"^2 - "x"^2)`= ω dt .....(3)
b. Integrating both sides of equation (3),
∫`"dx"/sqrt("A"^2 - "x"^2)` = ∫ω dt
∴ `sin^-1("x"/"A")` = ωt + Φ
where, α is constant of integration which depends upon initial condition (phase angle)
∴ `"x"/"A" = sin(ω"t" + Φ)`
∴ x = A sin(ωt + Φ)
This is the required expression for displacement of a particle performing linear S.H.M. at time t.
संबंधित प्रश्न
Answer in brief.
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