Advertisements
Advertisements
Question
Using the differential equation of linear S.H.M., obtain an expression for acceleration, velocity, and displacement of simple harmonic motion.
Advertisements
Solution
- 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.
RELATED QUESTIONS
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s2 to 9.8 m/s2.
Acceleration of a particle executing S.H.M. at its mean position.
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 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 ______.
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) ____________.
Which of the following represents the acceleration versus displacement graph of SHM?
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)
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.)
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 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 ______.
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 ______.
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 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 ______.
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 ______.
State the expression for the total energy of SHM in terms of acceleration.
Which one of the following is not a characteristics of SHM?
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 pendulum is performing simple harmonic motion. The acceleration of the bob is 20 cm s−2 at a distance of 5 cm from mean position. The time period of oscillation is ______.
