In: Physics
A 0.26 kg mass is attached to a light spring with a force constant of 36.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following.
(a) maximum speed of the oscillating mass
m/s
(b) speed of the oscillating mass when the spring is compressed 1.5
cm
m/s
(c) speed of the oscillating mass as it passes the point 1.5 cm
from the equilibrium position
m/s
(d) value of x at which the speed of the oscillating mass
is equal to one-half the maximum value
m
Mass of the object, m = 0.53 kg
Force constant, k = 23.2 N/m
Angular frequency, ? = ? (k/m) = 11.91 rad/s
= 6.62 rad/s
Solution:
(a)
Amplitude, A = 0.05 m
Maximum speed, vmax = ? A
= 11.91* 0.05
= 0.596m/s
(b)
Distance of compression, x = 0.015 m
Speed, v = ? ? [A^2 - x^2]
= 11.91 ? [0.05^2 - 0.015^2]
= 0.568 m/s
(c)
Distance of stretching, x = 0.015 m
Speed, v = ? ? [A^2 - x^2]
= 11.91 ? [0.05^2 - 0.015^2]
= 0.568 m/s
(d)
v = vmax / 2
? ? [A^2 - x^2 ] = ? A / 2
A^2 - x^2 = A^2 / 4
x^2 = 3 A^2 / 4
x = (?3 / 2) A
= 0.866 * 0.05
= 0.0433 m