A51.0 g stone is attached to the bottom of a vertical spring and set vibrating. if the maximum speed of the stone is 16.0 cm/s and the period is 0.250 s, find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.
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Ответ:
A) 32.22 N/m b) 0.0156 m c) 4 Hz
Explanation:
Using Hooke's law;
T = 2π √m/k where m is mass of the body in kg and k is the force constant of the spring N/m and T is the period of vibration in s.
M = 51 g = 51 / 1000 in kg = 0.051kg
Make k subject of the formula
T/2π = √m / k
Square both sides
T^2 / 4π^2 = m/k
Cross multiply
K = 4 π^2 * m/T^2
K = 4 * 3.142 * 3.142 * 0.051/ 0.25^2= 32.22N/m
B) using Hooke's law;
F = k e where e is the maximum displacement of the spring from equilibrium point called amplitude
F= weight of the body = mass * acceleration due to gravity = 0.051*9.81
0.5 = 32.22 * e
e = 0.5/32.22 = 0.0156 m
C) frequency is the number of cycle completed in a second = 1 / period
F = 1 / 0.25 = 4Hz
Ответ:
I hope this helps a lot.