At what position displacement velocity and acceleration of a particle performing SHM is maximum?

Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.

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Let’s obtain the differential equation for linear simple harmonic motion……..!
Now we shall derive the formulae for acceleration and velocity in linear SHM…….!
Some important values for acceleration and velocity:-
Let’s solve some numerical to understand in more details…..!

The differential equation of linear SHM is `(d^2vecx)/(dt^2) + k/m vecx = 0` where m = mass of the particle performing SHM.

(d^2vecx)/(dt)^2 = acceleration of the particle when its displacement from the mean position is `vecx` and k = force constant. For linear motion, we can write the differential equation in scalar form :

`(d^2x)/(dt^2) + k/m x = 0`

Let `k/m = omega^2` , a constant

`:. (d^2x)/(dt^2) + omega^2x = 0`

∴ Acceleration, a = `(d^2x)/(dt^2) = -omega^2x` ....(1).

The minus sign shows that the acceleration and the displacement have opposite directions. Writing
v = `(dx)/(dt)` as the velocity of the particle.

`a = (d^2x)/(dt^2) = (dv)/(dt) = (dv)/(dx)xx(dx)/(dt) = (dv)/(dx) = v = v (dv)/(dx)`

Hence, Eq. (1) can be written as

`v(dv)/(dx) = -omega^2x`

∴ `vdv = -omega^2x`

Integrating this expression, we get

`v^2/2 = (-omega^2x^2)/2 + C`

where the constant of integration C is found from a boundary condition.

At an extreme position (a turning point of the motion), the velocity of the particle is zero. Thus, v = 0 when x = ±A, where A is the amplitude.

`:. 0 = (-omega^2A^2)/2 + C` `:. C = (omega^2A^2)/2`

`:.v^2/2 = (-omega^2x^2)/2 + (omega^2A^2)/2`

`:. v^2 = omega^2(A^2 -x^2)`

`:.v = +- omegasqrt(A^2 - X^2)` ......(2)

This equation gives the velocity of the particle in terms of the displacement, x. The velocity towards right is taken to be positive and towards left as negative.

Since v = dx/dt we can write Eq. 2 as follows:

`(dx)/(dt) = omegasqrt(A^2 - x^2)`(considering only the plus sign)

`:. (dx)/(sqrt(A^2-x^2)) = omega dt`

Integrating the expression, we get,

`sin^(-1) (x/A) = omegat+x` ....(3)

where the constant of integration, x , is found from the initial conditions, i.e., the displacement and the
velocioty of the particle at time t = 0.

From Eq (3), we have

`x/A= sin(omegat + x)`

∴ Displacement as a function of time is, x = Asin (ωt + x).

We know that when a body is set up in vibrations from its original/mean position by external force it come backs to its mean position due to action of restring force present in the body and said to be performing the linear simple harmonic motion. Whereas the linear motion of object along the circumference of circle with uniform velocity is known as uniform circular motion. The acceleration of particle in linear SHM is given as,

Where ω is angular velocity and x is displacement of particle,given as

ω = 2π/T

(T is period of SHM)

Let’s obtain the differential equation for linear simple harmonic motion……..!

Consider a particle of mass m performing linear SHM with angular velocity ω. Let x be the displacement of particle from its mean position.

Then the restoring force acting on the particle is given as,

F= -Kx……..(1)

Also by Newton’s 2nd law of motion,

F = ma ……..(2)

From equations (1) and (2) we get,

ma = -kx……..(3)

Now the acceleration of the particle is given as,

a = dv/dt

but velocity, v= dx/dt

then acceleration can be written as,

Now we shall derive the formulae for acceleration and velocity in linear SHM…….!

1) Acceleration in SHM:-

We know the differential equation of SHM is,

From above equation we can conclude that acceleration in SHM is directly proportional to the displacement, and acts in opposite direction of displacement.

2) Velocity in SHM:-

Using equation for acceleration, we can find out the velocity of particle in SHM as follows,

Then equation (2) becomes,

This is an equation for velocity of particle performing linear SHM, where is angular frequency.

Some important values for acceleration and velocity:-

1.) Acceleration:-

When particle is at extreme position, x=A

∴ accelerationmax = – ω2 A

When particle is at mean position, x=0

∴ accelerationmin = 0

2) Velocity:-

When particle is at mean position, x=0

∴Vmax = Aω

When particle is at extreme position, x = 0

∴Vmin = 0

Let’s solve some numerical to understand in more details…..!

Ex:1) A particle performs SHM with period of 3π s. If amplitude of SHM is 12 cm, find the velocity and acceleration of particle at distance of 6 cm from mean position.

Solution:

Here, T = 3π s, A = 12 cm, x = 6 cm

Now we have,

ω = 2π/T

∴ω = 2π/3π

∴ω = 2/3 rad/s

1.) Acceleration is given as,

Answer

Verified

Ans: The amplitude of motion is 22.92 cm

Example – 7:

A particle in simple harmonic motion has a velocity of 10 cm/s when it crosses the mean position. If the amplitude of its oscillations is 2 cm, find the velocity. When it is midway between the mean and extreme positions.

Given: Velocity at mean position = vmax = 10 cm/s, amplitude = a = 2 cm, Displacement midway between the mean and extreme positions, hence x = a/2 = 2/2 = 1 cm.

To Find: Velocity = v =?

Solution:

We have vmax = ωa

∴ 10 = ω x 2

∴ ω = 10/2 = 5 rad/s

Ans: velocity at midway between the mean and extreme positions is 8.66 cm/s

Example – 8:

Show that the velocity of a particle performing simple harmonic motion is half the maximum velocity at a displacement of √3/2 times its amplitude.

Given: Displacement x = a√3/2

To Show: v = 1/2 vmax.

Solution:

Example – 9:

A particle performs S.H.M. of amplitude 10 cm. Its maximum velocity during oscillations is 100 cm/s. What is its displacement when the velocity is 60 cm/s?

Given: amplitude = 10 cm, Vmax = 100 cm/s, v = 60 cm/s

To Find: displacement = x =?

Solution:

vmax = ωa

∴ ω = vmax/a = 100/10 = 10 rad/s

Ans: Displacement = 8 cm

Example – 10:

A particle performing S.H.M. along a straight line has a velocity of 4π cm/s when its displacement is √12 cm. If the maximum acceleration it can attain is 16π2 cm/s2, find the amplitude and the period of its oscillations.

Given: vmax = 4π cm/s, f max = 16π2 m/s2 , Displacement = √12 cm

To Find: Amplitude = a =? and Period = T = ?

Solution:

fmax = ω2a

∴ 16π2 = ω2a ………. (2)

From equations (1) and (2) we have

ω2(a2 – 12) = ω2a

∴ (a2 – 12) = a

∴ a2 – 12 – a = 0

∴ (a – 4)(a + 3) = 0

∴ a = 4 cm or a = – 3 cm

Amplitude is maximum displacement hence a = 3 cm < √12 cm is not possible.

∴ a = 4 cm

substituting in equation (2)

16π2 = ω2(4)

∴ ω2 = 4π2

∴ ω = 2π rad/s

∴ T = 2π /ω = 2π /2π = 1 s

Ans: amplitude = 4 cm and period = 1 s

Example – 11

A particle of mass of 10 g performs S.H.M. of period 5 s and has an amplitude of 8 cm. Find its velocity when it is at a distance of 6 cm from the equilibrium position. Find also the maximum velocity and maximum force acting on it.

Given: mass = m = 10 g, Period = T = 5 s, amplitude = a = 8 cm, displacement = x = 6 cm, particle passes through mean position, α = 0.

To Find: velocity = v = ?, vmax = ?, Fmax = ?

Solution:

Angular velocity = ω = 2π/T = 2π/5 rad/s

∴ vmax = ωa = 2π/5 x 8 = 10.05 cm/s

fmax = ω2a = ( 2π/5)2 x 8 = 12.63 cm/s2

Fmax = m. fmax = 10 x 12.63 = 126.3 dyne

∴ Fmax = 126.3 x 10-5 N = 1.263 x 10-3 N

Ans: velocity =6 .65 cm/s; maximum velocity =10.05 cm/s; maximum force = 1.263 x 10-3 N

Example – 12:

If a particle performing S.H. M. starts from the extreme position after an elapse of what fraction of the period will the velocity of the particle be half the maximum velocity?

Given: v = 1/2 vmax. particle starts from extreme position, α = π/2.

Fo Find: Fraction of time = t/T =?

Solution:

∴ 4a2 – 4x2 = a2

∴ 4x2 = 3a2

∴ 2x = a√3

∴ x = a√3/2

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴ a√3/2 = 1 sin ((2π/T)t + π/2)

∴ 3/2 = cos ((2π/T)t)

∴ (2π/T)t = cos-1(3/2) = π/6

∴ t /T = 1/12 s

Ans: fraction of the period is 1/12 s

Example – 13:

A particle performs a linear S.H.M. Its velocity is 3 cm/s when it is at 4 cm from the mean position and 4 cm/s when it is at 3 cm from the mean position. Find the amplitude and the period of S.H.M.

Given: v1 = 3 cm/s at x1 = 4cm and v2 = 4 cm/s at x2 = 3cm

To Find: Amplitude = a =? Period = T=?

Solution:

∴ 16a2 -256 = 9a2 -81

∴ 16a2 – 9a2 = 256 – 81

∴ 7a2 = 175

∴ a2 = 25

∴ a = 5

Now T = 2π/ω = 2 x 3.14 /1 = 6. 28 s

Ans: Amplitude = 5 cm and period = 6.28 s

Example – 14

The velocities of a particle performing linear S.H.M. are 0.13 m/s and 0.12 m/s when it is at 0.12 m and 0.13 m respectively from the mean position. Find its period and amplitude.

Given: v1 = 0.13 m/s = 13 cm/s at x1 = 0.12 m = 12 cm and v2 = 0.12 m/s = 12 cm/s at x2 = 0.13 m = 13 cm

To Find: Amplitude = a =? Period = T=?

Solution:

∴ 144a2 – 144 x 144 = 169a2 – 169x 169

∴ 169a2 – 144a2 = 169 x 169 – 144x 144

∴ 25a2 = (169 + 144)(169 – 144)

∴ 25a2 = (313)(25)

∴ a2 = 313

∴ a = √313 = 17.69 m

Now T = 2π/ω = 2 x 3.14 /1 = 6. 28 s

Ans: Period =6.28 s and amplitude = 17.69 cm

Example – 15:

A particle performing S.H.M. has velocities of 8 cm/s and 6 cm/s at displacements of 3 cm and 4 cm respectively. Find its amplitude and frequency of oscillations. Calculate its maximum velocity. What is the phase of its motion when the displacement is 2.5 cm?

Solution:

Given: v1 = 8 cm/s at x1 =3 cm and v2 = 6 cm/s at x2 = 4 cm, displacement = x = 2.5 cm

To Find: Amplitude = a =? frequency = n = ?, phase = (ωt + α) =?,

∴ 9a2 – 81 = 16a2 – 256

∴ 16a2 – 9a2 = 256 – 81

∴ 7a2 = 175

∴ a2 = 25

∴ a = 5 cm

Now ω = 2 π n

∴ n = ω/2π = 2/( 2 x 3.142) = 0.3183 Hz

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)