Answer to Question #166942 in Quantum Mechanics for Kenan

Question

The current position of the harmonic oscillator is given by the relation 

x(t)=(0,4m)cos(πt/2).Time t is expressed in seconds. Find the maximum speed of the oscillator

Answer 1

Let's first find the speed of the oscillator by taking the derivative from $x(t)$ with respect to $t$:

$$v(t)=\dfrac{d}{dt}(x(t)),$$ $$v(t)=\dfrac{d}{dt}(0.4\ m\cdot cos(\dfrac{\pi t}{2}))=-0.4\ m\cdot\dfrac{\pi}{2}\ \dfrac{rad}{s}sin(\dfrac{\pi t}{2}),$$ $$v(t)=-0.63\ \dfrac{m}{s}\cdot sin(\dfrac{\pi t}{2}).$$

The maximum speed of the oscillator will be when $sin(\dfrac{\pi t}{2})=1$. Therefore, we get:

$$v_{max}=-0.63\ \dfrac{m}{s}.$$

The magnitude of the maximum velocity of the oscillator equals $0.63\ \dfrac{m}{s}$.