A 0.200-kg mass is attached to the end of a spring with a spring constant of 11
ID: 1786251 • Letter: A
Question
A 0.200-kg mass is attached to the end of a spring with a spring constant of 11 N/m. The mass is first examined (t = 0) when the mass is 17.0 cm from equilibrium in the positive x-direction, and is traveling at 2.0 m/s in the positive x-direction.
a) Write an equation x(t) that describes the position of this mass as a function of time. Express this function in terms of numerical values, trigonometric functions and the time variable “t”.
b) Repeat for v(t), the speed of the mass as a function of time.
c) Repeat for a(t), the acceleration of the mass as a function of time
Explanation / Answer
given m = 0.2 kg
k = 11 N/m
at x = 17 cm = 0.17 m
v = 2.0 m/s
a) angular frequency of motion, w = sqrt(k/m)
= sqrt(11/0.2)
= 7.42 rad/s
let A is the amplitude.
using conservation of energy
(1/2)*k*A^2 = (1/2)*k*x^2 + (1/2)*m*v^2
(1/2)*11*A^2 = (1/2)*11*0.17^2 + (1/2)*0.2*2^2
==> A = 0.319 m
let phi is initial phase
at t = 0,
x = A*sin(phi)
phi = sin^-1(x/A)
= sin^-1(0.17/0.319)
= 0.562 rad
a) so, x(t) = A*sin(w*t + phi)
= 0.319*sin(7.42*t + 0.562) <<<<<<<<----------Answer
b) v(t) = dx/dt
= 0.319*cos(7.42*t + 0.562)*7.42
= 2.37*cos(7.42*t + 0.562) <<<<<<<<----------Answer
c) a(t) = dv/dt
= 2.37*(-sin(7.42*t + 0.562))*7.42
= -17.6*sin(7.42 + 0.562) <<<<<<<<----------Answer
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