A simple harmonic oscillator consists of a block ofmass 4.20 kg attached to aspr
ID: 1675583 • Letter: A
Question
A simple harmonic oscillator consists of a block ofmass 4.20 kg attached to aspring of spring constant 194 N/m.When t = 3.00 s, the position and velocity of the blockare x =+0.129 m and v = +3.415 m/s. (a) What is the amplitude of the oscillations?m
(b) What was the position of the mass at t = 0s?
x = m
(c) What was the velocity of the mass at t = 0s?
v = m/s (a) What is the amplitude of the oscillations?
m
(b) What was the position of the mass at t = 0s?
x = m
(c) What was the velocity of the mass at t = 0s?
v = m/s
Explanation / Answer
Equation of motion for simple harmonicmotion is x = A *sin ( * t + ) &velocity v = * A * cos ( *t + ) = *(A2 - x2) A = Amplitude = angularfrequency = (k/m) = (194/ 4.20) = 6.796 rad/s = initialphase given at t = 3.00 s, x3.0 = +0.129 m, v3.0 = +3.415 m/s a. v3.0 = *(A2 - x3.02) 3.4152 = 6.7962*(A2 - 0.1292) A2 = 11.662/ 46.186 + 0.0166 Amplitude A = 0.2691 = 0.519 m b. x3.0 = A* sin ( * t + ) 0.129 = 0.519* sin (6.796 * 3.0 + ) sin (20.388 + ) = 0.129 /0.519 initialphase = sin-1(0.129 / 0.519) - 20.38 = 0.251 - 20.388 = - 20.137 rad x = A* sin ( *t - 20.137) at t = 0 x0 = 0.519* sin ( *0 - 20.137) = -0.498 m c. v0 = 6.796* 0.519 * cos ( *0 - 20.137) = 0.986 m/sRelated Questions
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