A charged particle with a mass of 0.01 kg and a charge of 1 C oscillates on a spring with a frequency of 100 Hz in the presence of a magnetic field of 0.5 T perpendicular to the motion. Calculate the amplitude of oscillation.
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The frequency of oscillation, \(f\), is related to the angular frequency, \(\omega\), by the equation:
\[f = \frac{\omega}{2\pi}\]
The angular frequency, \(\omega\), can be found using the equation for the period of oscillation, \(T\):
\[T = \frac{1}{f} = \frac{2\pi}{\omega}\]
The period of oscillation, \(T\), is the reciprocal of the frequency:
\[T = \frac{1}{100 \, \text{Hz}} = 0.01 \, \text{s}\]
The angular frequency, \(\omega\), is then:
\[\omega = \frac{2\pi}{0.01 \, \text{s}} = 628.32 \, \text{rad/s}\]
The Lorentz force on a charged particle moving in a magnetic field is given by the equation:
\[F = qvB\]
The force on the particle must be equal to the force exerted by the spring for the particle to oscillate. The force exerted by the spring can be given by Hooke's Law:
\[F = -kx\]
where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. Equating these two forces gives:
\[qvB = -kx\]
Since the force on the particle is proportional to the acceleration of the particle, we can use Newton's second law, \(F = ma\), to express the equation above in terms of acceleration:
\[qvB = ma\]
Since acceleration is the second derivative of displacement with respect to time, we have:
\[qvB = m\frac{{d^2x}}{{dt^2}}\]
Given that the mass of the particle is \(m = 0.01 \, \text{kg}\), the charge of the particle is \(q = 1 \, \text{C}\), and the magnetic field is \(B = 0.5 \, \text{T}\), we can rewrite the equation as:
\[0.01 \times 1 \times 0.5 = 0.01 \times \frac{{d^2x}}{{dt^2}}\]
Simplifying the equation gives:
\[0.005 = \frac{{d^2x}}{{dt^2}}\]
The general solution to this differential equation is:
\[x = A\sin(\omega t + \phi)\]
where \(x\) is the displacement from the equilibrium position, \(A\) is the amplitude of oscillation, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the phase constant. Since we are given that the frequency of oscillation is \(100 \, \text{Hz}\), we can write the equation as:
\[x = A\sin(2\pi(100)t + \phi)\]
Since a full cycle of oscillation is completed in one period, the time taken for one cycle is \(T = \frac{1}{f} = 0.01 \, \text{s}\). At \(t = T\), the displacement \(x\) must return to its initial value. This implies that \(\sin(2\pi(100)(T) + \phi) = \sin(\phi) = \sin(0) = 0\).
Hence, \(\phi = n\pi\) for integer values of \(n\). In other words, the phase constant is \(0\), \(\pi\), \(2\pi\), \(-\pi\), etc.
To determine the amplitude of oscillation, we need to know the initial conditions of the system, such as the initial displacement and velocity of the particle. Without these initial conditions, we cannot determine the specific value of the amplitude.
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