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 loop penetrate 4 cm into the magnetic field.
Explanation:
Given that,
Width w= 5 cm
Length L= 10 cm
mass m = 40 g
Resistance R = 20 mΩ
Initial velocity = 1 m/s
Magnetic field = 2 T
We need to calculate the induced emf
Using formula of emf
Put the value into the formula
We need to calculate the current
Using Lenz's formula
We need to calculate the force
Using formula of force
Put the value into the formula
We need to calculate the acceleration
Using formula of acceleration
Put the value in to the formula
We need to calculate the distance
Using equation of motion
Hence, The loop penetrate 4 cm into the magnetic field.