Answer to Question #276401 in Physics for Rakibul

Question

Define the simple harmonic motion (SHM).

(b) Write down the difference between Periodic and SHM?

(c) Derive the differential equation of SHM. A body is in simple harmonic motion is

having an amplitude of 5 cm and a period of 0.2 s. Calculate the acceleration and

the velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

Answer 1

(a) Harmonic oscillations are periodic oscillations of a physical quantity (or any other) depending on time, which occur according to the laws of sine or cosine.

(b) Periodic oscillations are those in which all the characteristics of motion are repeated after a certain period of time. The difference between (SHM) and periodic motion is that periodic motion refers to any type of periodic motion, while (SHM) refers to a specific type of periodic motion, where the restoring force is proportional to the displacement.

(c) $\frac{d^2x}{dx^2}+\omega_0^2x=0$

$x(t)=A\sin(\omega_0t+\phi)$ , $\omega_0=2\pi/T=2\cdot3.14/0.2=31.4\ (rad/s)$ . Assume that $\phi=0.$

So, we get

$x(t)=0.05\cdot\sin(31.4t)$

$v(t)=dx/dt=1.57\cdot\cos(31.4t)$

$a(t)=dv/dt=-49.298\cdot\sin(31.4t)$

If $x(t)=0.05\ (m)$ then $t=0.05\ (s)$

If $x(t)=0.03\ (m)$ then $t=0.02\ (s)$

If $x(t)=0\ (m)$ then $t=0\ (s)$

$v(0.05)=0.00125\ (m/s)$ , $v(0.02)=1.27\ (m/s)$ , $v(0)=1.57\ (m/s)$

$a(0.05)=-49.3\ (m/s^2)$ , $a(0.02)=-28.96\ (m/s^2)$ , $a(0)=0\ (m/s^2)$