Question

show that when a particle is moving in a SHM,its velocity at a distance of sqrt of 3/2A from the central position is half of its velocity in the central position

Answer 1

Show that when a particle is moving in a SHM, its velocity at a distance of $x_{1} = \sqrt{3 / 2} * A$ from the central position is half of its velocity in the central position.

Solution

$$m a = - k x \iff \ddot {x} + \frac {k}{m} x = 0$$ $$x ^ {\prime \prime} = \frac {d v}{d t}$$ $$v = \int - \frac {k}{m} * x d x$$

$v_{1} = \int_{A}^{\sqrt{3 / 2} *A} - \frac{k}{m} * x dx = -\frac{k}{m} * \left(\frac{3 / 2*A^{2}}{2} -\frac{A^{2}}{2}\right) = \frac{k*A}{2} * \left(-\frac{3}{2} +1\right) = \frac{k*A}{2} * \left(-\frac{1}{2}\right)$ - velocity at a distance of $x_{1} = \sqrt{3 / 2}$

$v_{2} = \int_{A}^{0} - \frac{k}{m} * x dx = -\frac{k}{m} * \left(0 - \frac{A^{2}}{2}\right) = \frac{k*A}{2} * (0 + 1) = \frac{k*A}{2} * 1$ velocity at a distance of $x_{0} = 0$

$$\frac {v _ {1}}{v _ {2}} = - \frac {1}{2}$$