Answer to Question #102596 in Mechanics | Relativity for Ojugbele Daniel

Question

The position of a simple harmonic oscillator as a function of time is given by x=4.6cos(7πt/6 + π/8) where t is in second and x is in meter. Find: (i) the frequency (ii) the velocity and acceleration at t=0.

Answer 1

As per the question,

The equation of the wave $x=4.6\cos(\dfrac{7πt}{6} + π/8)$

Now comparing this the general SHM equation, $X=A cos(\omega t+\phi)$

a)

$\omega=\dfrac{7\pi}{6}$

$\Rightarrow 2\pi f=\dfrac{7\pi}{6}$

$\Rightarrow f=\dfrac{7\pi}{6\times2\pi}=\dfrac{7}{12}Hz$

b)

$x=4.6\cos(\dfrac{7πt}{6} + π/8)$

$v=\dfrac{dx}{dt}=-4.6\times \dfrac{7\pi}{6}\sin(\dfrac{7\pi t}{6}+\dfrac{\pi}{8})$

at t=0

$v=-4.6\times \dfrac{7\pi}{6}\sin(\dfrac{\pi}{8})=-11.91m/sec$

$a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}=-4.6\times \dfrac{7\pi}{6}\times \dfrac{7\pi}{6}\cos(\dfrac{7\pi t}{6}+\dfrac{\pi}{8})$

at t=0,

$a=-4.6\times \dfrac{7\pi}{6}\times \dfrac{7\pi}{6}\sin(\dfrac{\pi}{8})=-\dfrac{225.4\pi^2}{36}\times\dfrac{1}{\sqrt{2}}$

$a=-43.73m/sec^2$