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Answer:
If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2
Explanation:
We know that in a simple harmonic oscillator the maximum speed is given by
=
Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to of the oscillation .
Since
= 2
Where is the maximum speed when frequency is doubled .
Answer:
1.92 x 10⁻¹²J
Explanation:
The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e
F = ma
Where;
a = [v = linear velocity, r = radius of circular path]
=> F = m ------------(i)
We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;
F = q v B sin θ ----------(ii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = the angle between the velocity and the magnetic field.
Combining equations (i) and (ii) gives
m = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]
=> m = q v B sin 90°
=> m = q v B
Divide both side by v;
=> m = qB
Make v subject of the formula
v =
From the question;
B = 1.25T
m = mass of proton = 1.67 x 10⁻²⁷kg
r = 0.40m
q = charge of a proton = 1.6 x 10⁻¹⁹C
Substitute these values into equation(iii) as follows;
v =
v = 4.79 x 10⁷m/s
Now, the kinetic energy, K, is given by;
K = mv²
m = mass of proton
v = velocity of the proton as calculated above
K =
K = 1.92 x 10⁻¹²J
The kinetic energy is 1.92 x 10⁻¹²J