Question

Using differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.​

Answer 1

Let
`A`,
`B`, and
`k` denote three constants (with the requirement that
`k > 0`.) The following assumes that the mass of this object is
`m`. Assume that
`x(t)` denotes the position of the object at time
`t`.

(a)
`\displaystyle x^\prime(t) = -\sqrt{(k)/(m)}\, A\, \sin\left(t\, \sqrt{(k)/(m)}\right) + \sqrt{(k)/(m)}\, B\, \cos\left(t\, \sqrt{(k)/(m)}\right)`.

(b)
`\displaystyle x^(\prime\prime)(t) = -\left((k)/(m)\right)\, A\, \cos\left(t\, \sqrt{(k)/(m)}\right) - \left((k)/(m)\right)^(2)\, B\, \sin\left(t\, \sqrt{(k)/(m)}\right)`

Step-by-step explanation:

The differential equation for a simple harmonic motion might take the following form:

`\displaystyle (d^(2) x)/(d t^(2)) = -(k)/(m)\, x`.

The minus sign on the right-hand side highlights the fact that the displacement and acceleration of the object should be in opposite directions.

Notice how this equation is in the form of a homogeneous second-order ODE:

`x^(\prime\prime)(t) + \underbrace{P(t)}_(0)\, x(t) + \underbrace{Q(x)}_(√(k / m)) = 0`

Let
`r` be a constant. One possible solution to this homogeneous second-order ODE would be in the form
`x(t) = e^(r\, t)`, such that
`x^(\prime)(t) = r\, e^(r\, t)` whereas
`x^(\prime\prime)(t) = r^(2)\, e^(r\, t)`.

Substitute into the original ODE to obtain:

`\displaystyle \underbrace{r^(2)\, e^(r t)}_(x^(\prime\prime)(t)) + (k)/(m)\, \underbrace{e^(r t)}_(x(t)) = 0`.

Rearrange the equation and solve for
`r`.

`\displaystyle e^(r t)\, \left(r^(2) + (k)/(m)\right) = 0`

Notice that
`e^(r\, t) > 0`. Hence, it must be true that
`\displaystyle r^(2) + (k)/(m) = 0`. Solve for
`r` given that
`k > 0`:

`\displaystyle r_(1, 2) = \pm i\sqrt{(k)/(m)}`, where
`i` is the imaginary unit.

The two particular solutions for the ODE would be:

`x_1(t) = e^(\left(i\,√(k/m)\right)\, t)` and
`x_2(t) = e^(\left(-i\,√(k/m)\right)\, t)`.

Apply Euler's Formula to rewrite both solutions in terms of trigonometric functions:

`\displaystyle x_1(t) = e^(\left(i\,√(k/m)\right)\, t)= \sqrt{(k)/(m)} \left(\cos\left( \sqrt{(k)/(m)} t\right) + + i\, \sin\left( \sqrt{(k)/(m)} t\right)\right)`.

`\displaystyle x_2(t) = e^(\left(-i\,√(k/m)\right)\, t)= -\sqrt{(k)/(m)} \left(\cos\left( -\sqrt{(k)/(m)} t\right) + i\, \sin\left( -\sqrt{(k)/(m)} t\right)\right)`.

The general solution would be in the form:

`\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{(k)/(m)} t\right) + B\, \left(i\, \sin\left(\sqrt{(k)/(m)} t\right)\right)\right)`,

Where
`C_1` and
`C_2` are constants (not necessarily real numbers.)

Since position is supposed to assume a real value for any real
`t`, set
`B` to a multiple of
`i` such that the general solution is real-valued:

`\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{(k)/(m)} t\right) + B\, \sin\left(\sqrt{(k)/(m)} t\right)\right)`.

Differentiate to obtain general expressions for velocity (first derivative) and acceleration (second derivative.)