Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
The rate at which the height is changing is ( 5 / x ) m / hr
We know that,
Area of an equilateral triangle A =
/ 4
h = x / 2
Where,
x = Side
h = Height
Given that,
dA / dt = 5 / hr
h = x / 2
Differentiate both sides with respect to t
dh / dt = ( / 2 ) ( dx / dt )
dx / dt = ( 2 / ) ( dh / dt )
A =
/ 4
Differentiate both sides with respect to t
dA / dt = ( / 4 ) ( 2x ) ( dx / dt )
5 = ( / 4 ) ( 2x ) ( 2 /
) ( dh / dt )
dh / dt = ( 5 / x ) m / hr
Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt
Therefore, the rate at which the height is changing is ( 5 / x ) m / hr
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