You have an oscillating 0.5kg mass attached to a horizontal spring whose motion

Question

You have an oscillating 0.5kg mass attached to a horizontal spring whose motion can be described by x(t) = 20cos(8pit), where x is in mm and t is in seconds. You want to know how fast it is moving when x = 10 mm?

A. Find a solution while using v(t) = dx(t)/dt and no energy.

B. Find a solution while using E = K + U, but not v(t) = dx(t)/ dt.

C. Now chose your method of solution: You have a 0.25 kg block udergoing SHM on the end of a horizontal spring with spring constant k = 250 N/m. When the block is 0.025 m from its equilibrium point you measure the velocity to be 0.5 m/s. What is x(t) ?

Explanation / Answer

A.
x(t) = 20cos(8pit)
at x= 10 mm calculate time , t
x(t) = 20cos(8pit)
10=20cos(8pit)
8pi*t= pi/3
t= 1/24 s

V(t) = dx(t)/dt
= - 20*8pi* sin (8pi*t)
at t = 1/24 s
V= 20*8pi*sin(8*pi/24)
=453.3 mm/s
= 0.4533 m/s
Answer: 0.4533 m/s

B)
angular frequancy,w = 8pi
w= 2pi*sqrt(m/k)
8pi= 2pi*sqrt(0.5/k)
0.5/k=16
k=0.03125 N/m

Total energy E = 0.5*K*A^2 = 0.5*0.03125*(0.02)^2 = 6.25*10^-6 J {A= 20mm = 0.02 m}
at x= 10mm = 0.01 m
U = 0.5*k*x^2 = 0.5*0.03125*(0.01)^2 = 1.5625*10^-6 J

K= 0.5*m*v^2

use:
E = K + U
6.25*10^-6 = 0.5*m*v^2 + 1.5625*10^-6
0.5*0.5*v^2 = 4.6875*10^-6
v=0.00433 m/s
Answer: 0.00433 m/s

C)
Isn't x(t) suppose to be distance from equilibrium position
x(t) = 0.025 m

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