Question

The displacement of a harmonic oscillator is given by x(t)=9.4 sin (15t), where the unit of x are

meters and t is measures in seconds.

(a) Find the amplitude and the frequency. (b) What is the maximum velocity of the motion?

(c) What is the maximum acceleration?

Answer 1

(a) The amplitude is $A = 9.4\ m$.

We can find the frequency from the formula:

$$\omega=2\pi f,$$ $$f=\dfrac{\omega}{2\pi}=\dfrac{15\ \dfrac{rad}{s}}{2\pi}=2.38\ Hz.$$

(b) Let’s take the derivative of $x(t)$ with respect to time:

$$v(t)=\dfrac{d}{dt}(9.4sin(15t))=9.4\ m\cdot15\ \dfrac{rad}{s}\cdot cos(15t),$$ $$v(t)=141\ \dfrac{m}{s}\cdot cos(15t).$$

We can obtain the maximum velocity of the motion when $cos(\omega t)=1$, therefore:

$$v(t)_{max}=141\ \dfrac{m}{s}.$$

(c) Let’s take the derivative of $v(t)$ with respect to time:

$$a(t)=\dfrac{d}{dt}(141\ \dfrac{m}{s}\cdot cos(15t))=-141\ \dfrac{m}{s}\cdot15\ \dfrac{rad}{s}sin(15t),$$ $$a(t)=-2115\ \dfrac{m}{s^2}\cdot sin(15t).$$

We can obtain the maximum acceleration of the motion when $sin(\omega t)=1$, therefore:

$$a(t)_{max}=-2115\ \dfrac{m}{s^2}.$$

Answer:

(a) $A = 9.4\ m, f=2.38\ Hz.$

(b) $v(t)_{max}=141\ \dfrac{m}{s}.$

(c) $a(t)_{max}=-2115\ \dfrac{m}{s^2}.$