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.
Answer:
The current of the outer coil is
Explanation:
From the question we are told that
The number of turns of the inner coil is
The radius of the inner coil is
The current of the inner coil is
The number of turns of the outer coil is
The radius of the outer coil is
Generally the net magnetic field is mathematically represented as
Now from told that the net magnetic field is common
So
Here is the permeability of free space
making the subject
substituting values