Question #102828

A 2kg body panel of a car oscillate with a frequency of 2Hz and magnitude of 2.5cm. if the oscillation is assume to be a simple harmonic oscillation and undamped calculate (i) maximum velocity of the panel. (ii) Total energy of the panel. (iii) The maximum potential of the panel. (iv) The Kinetic energy of the panel 1.0cm from it's equilibrium.

Expert's answer

Mass of the body panel of car=2kg

Frequency of oscillation=2Hz

So, angular velocity (ω\omega )=2πf=2π×2=4π2\pi f=2\pi\times2=4\pi

Amplitude =2.5cm=0.025m

Phase difference (ϕ)=0(\phi)=0

As per the question, oscillation is SHM, and undampped

The general equation of the SHM wave

y=Asin(ωt+ϕ)y=A \sin(\omega t+\phi)

y=0.025sin(4πt)y=0.025\sin(4\pi t)

a)

We know that in the case of maximum velocity(v)dydt=0.025×4πcos(4πt)\dfrac{dy}{dt}=0.025\times4\pi\cos(4\pi t)

at t=0, cos0=1\cos0^\circ=1

So, v=4×0.025×π=0.314m/secv=4\times0.025\times \pi=0.314m/sec

b)

Total energy of the system will be =KE+PEKE+PE

=mAω22+0=\dfrac{mA\omega^2}{2}+0 =2×0.025×4π×0.025×4π2=0.0986J\dfrac{2\times0.025\times 4\pi\times0.025\times 4\pi}{2}=0.0986J

c)

We know that, as per the energy conservation rule, total energy will be constant,

So, in case of maximum stored potential energy, KE=0

Hence, maximum potential energy so, PEmax=0.0986JPE_{max}=0.0986J

d)

Velocity, at the amplitude (x)=1cm

v=ωA2x2=4π0.02520.01=22.91×103m/secv=\omega\sqrt{A^2-x^2}=4\pi\sqrt{0.025^2-0.01}=22.91\times10^{-3}m/sec

The kinetic of the panel at the required position =mv22=2×22.91×103×22.91×1032\dfrac{mv^2}{2}=\dfrac{2\times 22.91\times 10^{-3}\times 22.91\times 10^{-3}}{2}

=524.8681×106J=5.25×106J=524.8681\times10^{-6}J=5.25\times10^{-6}J


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