A simple harmonic oscillator consists of a block of mass 1.83 kg attached to a s

Question

A simple harmonic oscillator consists of a block of mass 1.83 kg attached to a spring of spring constant 100 N/m. When t = 1.00 s, the position and velocity of the block are x = 0.122 m and v = 3.413 m/s. What is the amplitude of the oscillations? What was the position of the block at t = 0 s? What was the velocity of the block at t = 0 s?

Explanation / Answer

A simple harmonic oscillator consists of a block of mass 4.50 kg attached to a spring of spring constant 330 N? When t = 2.20 s, the position and velocity of the block are x = 0.189 m and v = 3.680 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 ANS:? = sqrt(k/m) = sqrt(330/4.5) = 8.56349 From SHM we have x = A*cos(?t + f) and v = -?A*sin(?t + f ) So dividing the 2nd eqn by the first gives v/x = -?*sin(?t + f)/cos(?t + f) = -?*tan(?t + f) so (?t + f) = -arctan(v/(x*?)) = - arctan(3.680/(0.189*8.56349)) = -1.15645 So f = -1.15645 - 8.56349*2.20 = -19.996 rad Now plugging into the position eqn we have x = A*cos(?t + f) => A = x/cos(?t + f) = 0.189/cos(8.56349*2.20 - 19.996) = 0.469m b) At t = 0 then x = 0.469*cos(0 - 19.996) = 0.193m c) v = -8.56349*0.469*sin(0 - 19.996) = -3.66m/s

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