Equations of Simple Harmonic Motion (basic) PLEASE! show work and only answer if
ID: 1466757 • Letter: E
Question
Equations of Simple Harmonic Motion (basic)
PLEASE! show work and only answer if you know how to do it. People keeps giving me the wrong answer.
Analyzing Newton's 2^nd Law for a mass spring system, we found a_x = -k/m X. Comparing this to the x-component of uniform circular motion, we found as a possible solution for the above equation: x = Acos(omega t) v_x = - omega Asin(omega t) a_x = - omega^2 Acos(omega t) with omega = square root k/m, and A the amplitude (maximum displacement from equilibrium). Consider an oscillator with mass 0.217 kg and spring constant 21.6 N/m. If the amplitude of the oscillation is 4.53 cm, what is the maximum speed? Consider an oscillator with frequency 31 Hz. If at t=0 the oscillator is at a maximum displacement from the equilibrium of +5.8 cm. what is the displacement 3.46 seconds later? Consider an oscillator with mass 0.31 kg and spring constant 14.2 N/kg. The oscillator is displaced from the equilibrium position by +5.9 cm and released. After what time, is the oscillator for the first time at a displacement of -2.25 cm?Explanation / Answer
Qx = -k/m x
solving the above condition we get,
x = A cos wt
v = - wA sin wt
a = - w2A cos wt
a) k =21.6 N/m , m= 0.217 kg A = 4.53 cm
v = -wA sin wt
for maximum speed, sin wt = -1
in the third quadrant, wt= 3pi/2 sin 3pi/2 = - sin pi/2 =-1
vmax = wA = sqrt(k/m) A = sqrt( 21.6/0.217) 4.53 x10-2
vmax = 0.45 m/s
b) f = 31 Hz, at t =0 , x = +5.8 cm
w = 2pi f = 2pi x 31 = 62 pi
x = A coswt
at t =0 5.8 = A cos 0
A = 5.8 cm
at t = 3.46 s, x = 5.8 cos ( 2pi x f x 3.46)
x = + 4.02 cm
c) m = 0.31 kg , k = 14.2 N/m
w = sqrt (m/k) = sqrt(0.31 /14.2) = 0.148
t =0 , x = +5.9 cm
x = A coswt
A = 5.9 cm
x = A cos wt
cos wt = x/A
cos wt = -2.25/5.9
= - 0.38
cos wt = cos 112 = cos (0.622 pi)
w t = 0.622 pi
t = 0.622 pi /0.148
t = 4.2 x3.14 s
t = 13.18 s
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