Answer to Question #199605 in Physics for Rakesh Kumar

Question

The equation of a particle executing SHM is x = 3 sin (2πt+π/6), where x is in meter and t is in seconds. 

Find the amplitude, time period, maximum speed and maximum acceleration. Also find the velocity and 

acceleration at t = 0.5 s.

Answer 1

a) $A=3\ m$.

b)

$$\omega=2\pi f=\dfrac{2\pi}{T},$$ $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2\pi s^{-1}}=1\ s.$$

c)

$$v=\dfrac{dx}{dt}=\dfrac{d}{dt}(3 sin(2\pi t+\dfrac{\pi}{6}))=3\cdot2\pi\ s^{-1} cos(2\pi t+\dfrac{\pi}{6}),$$ $$v=18.85\ \dfrac{m}{s}\cdot cos(2\pi t+\dfrac{\pi}{6}).$$

Maximum speed will be when $cos(2\pi t+\dfrac{\pi}{6})=1$:

$$v_{max}=18.85\ \dfrac{m}{s}.$$

d)

$$a=\dfrac{dv}{dt}=\dfrac{d}{dt}(18.85\ \dfrac{m}{s}\cdot cos(2\pi t+\dfrac{\pi}{6})),$$ $$a=-18.85\ \dfrac{m}{s}\cdot2\pi\ s^{-1}\cdot sin(2\pi t+\dfrac{\pi}{6})=-118.4\ \dfrac{m}{s^2}\cdot sin(2\pi t+\dfrac{\pi}{6}).$$

Maximum acceleration will be when $sin(2\pi t+\dfrac{\pi}{6})=1$:

$$a_{max}=-118.4\ \dfrac{m}{s^2}.$$

e)

$$v(t=0.5\ s)=18.85\ \dfrac{m}{s}\cdot cos(2\pi\ s^{-1}\cdot0.5\ s+\dfrac{\pi}{6}),$$ $$v(t=0.5\ s)=-16.32\ \dfrac{m}{s}.$$

f)

$$a(t=0.5\ s)=-118.4\ \dfrac{m}{s^2}\cdot sin(2\pi\ s^{-1}\cdot0.5\ s+\dfrac{\pi}{6}),$$ $$a(t=0.5\ s)=59.2\ \dfrac{m}{s^2}.$$