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

Question #102596
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.
1
Expert's answer
2020-02-10T09:27:04-0500

As per the question,

The equation of the wave x=4.6cos(7πt6+π/8)x=4.6\cos(\dfrac{7πt}{6} + π/8)

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


a)

ω=7π6\omega=\dfrac{7\pi}{6}

2πf=7π6\Rightarrow 2\pi f=\dfrac{7\pi}{6}

f=7π6×2π=712Hz\Rightarrow f=\dfrac{7\pi}{6\times2\pi}=\dfrac{7}{12}Hz


b)


x=4.6cos(7πt6+π/8)x=4.6\cos(\dfrac{7πt}{6} + π/8)

v=dxdt=4.6×7π6sin(7πt6+π8)v=\dfrac{dx}{dt}=-4.6\times \dfrac{7\pi}{6}\sin(\dfrac{7\pi t}{6}+\dfrac{\pi}{8})

at t=0

v=4.6×7π6sin(π8)=11.91m/secv=-4.6\times \dfrac{7\pi}{6}\sin(\dfrac{\pi}{8})=-11.91m/sec


a=dvdt=d2xdt2=4.6×7π6×7π6cos(7πt6+π8)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=4.6×7π6×7π6sin(π8)=225.4π236×12a=-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/sec2a=-43.73m/sec^2



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