To determine when velocity is maximum in Simple Harmonic Motion (SHM), we need to recall the equations for displacement, velocity, and acceleration in SHM and analyze their behavior at different points in the oscillation cycle.
Displacement in SHM: The displacement \(x\) of a particle undergoing SHM can be represented as:
\[ x(t) = A \sin(\omega t + \phi) \]where \(A\) is the amplitude, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the initial phase.
Velocity in SHM: The velocity \(v\) is the time derivative of displacement:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]The magnitude of velocity is maximum when \(|\cos(\omega t + \phi)|\) is maximum. The maximum value of \(|\cos(\theta)|\) is 1. Therefore, the maximum velocity \(v_{max}\) is:
\[ v_{max} = A \omega \]Condition for maximum velocity: For \(v\) to be maximum, \(\cos(\omega t + \phi)\) must be \(\pm 1\). This occurs when the phase \((\omega t + \phi)\) is \(0, \pi, 2\pi, \ldots\) or \(n\pi\) where \(n\) is an integer. When \(\cos(\omega t + \phi) = \pm 1\), then \(\sin(\omega t + \phi) = 0\).
Relating maximum velocity to displacement: Since \(x(t) = A \sin(\omega t + \phi)\), if \(\sin(\omega t + \phi) = 0\), then \(x(t) = A \times 0 = 0\). This means the displacement is zero.
Relating maximum velocity to acceleration: The acceleration \(a\) in SHM is the time derivative of velocity:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) \]When velocity is maximum, \(\sin(\omega t + \phi) = 0\), which implies \(a(t) = 0\). So, acceleration is zero when velocity is maximum.
Therefore, the velocity in SHM is maximum when the displacement is zero, i.e., at the equilibrium position.
C) Displacement is zero: As derived, velocity is maximum when \(\cos(\omega t + \phi) = \pm 1\), which corresponds to \(\sin(\omega t + \phi) = 0\). Since \(x = A \sin(\omega t + \phi)\), this means \(x=0\).
To determine the phase difference between displacement and velocity in Simple Harmonic Motion (SHM), we need to express both quantities mathematically and then compare their phase angles. Displacement in SHM is typically represented by a sinusoidal function, and velocity is the time derivative of displacement.
Displacement in SHM: The displacement of a particle undergoing SHM can be represented by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]where \(x(t)\) is the displacement at time \(t\), \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the initial phase angle.
Velocity in SHM: Velocity is the time derivative of displacement. Differentiating the displacement equation with respect to time:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt} [A \sin(\omega t + \phi)] \] \[ v(t) = A \omega \cos(\omega t + \phi) \]Convert cosine to sine: To compare the phases, it's easier if both functions are in the same trigonometric form (e.g., sine). We know that \(\cos(\theta) = \sin(\theta + \pi/2)\). Applying this identity to the velocity equation:
\[ v(t) = A \omega \sin(\omega t + \phi + \pi/2) \]Compare phases: Now we have the displacement and velocity equations in a comparable form:
The phase of displacement is \((\omega t + \phi)\).
The phase of velocity is \((\omega t + \phi + \pi/2)\).
Calculate phase difference: The phase difference is the difference between the phase of velocity and the phase of displacement:
\[ \text{Phase difference} = (\omega t + \phi + \pi/2) - (\omega t + \phi) \] \[ \text{Phase difference} = \pi/2 \]This means that velocity leads displacement by a phase of \(\pi/2\) radians, or 90 degrees.
D) The phase difference between displacement and velocity in SHM is \(\pi/
To determine the phase difference between displacement and acceleration in Simple Harmonic Motion (SHM), we need to recall their standard mathematical expressions and compare their arguments (phases).
Correct Option: A) The phase difference between displacement and acceleration in SHM is \(\pi\).
The question asks for the relationship between the force constant (\(k\)), mass (\(m\)), and angular frequency (\(\omega\)) in Simple Harmonic Motion (SHM). We need to recall the fundamental equations governing SHM, specifically the relationship between angular frequency, mass, and the force constant.
B) \(k = m\omega^2\) is the correct relationship derived from the principles of Simple Harmonic Motion.
The problem asks for the time period of a particle undergoing simple harmonic motion (SHM). The key to solving this is to recall the fundamental formula for the time period of an oscillator, which relates the mass of the oscillating body and the force constant (or spring constant) of the system.
Understand Simple Harmonic Motion (SHM): A particle undergoes SHM when the restoring force acting on it is directly proportional to its displacement from the equilibrium position and acts in the opposite direction. This is described by Hooke's Law for a spring-mass system: \(F = -kx\), where \(F\) is the restoring force, \(k\) is the force constant, and \(x\) is the displacement.
Relate Force to Acceleration: According to Newton's second law, \(F = ma\), where \(m\) is the mass of the particle and \(a\) is its acceleration. So, for SHM, we have \(ma = -kx\).
Define Angular Frequency: The acceleration in SHM is also given by \(a = -\omega^2 x\), where \(\omega\) is the angular frequency. Substituting this into the equation from step 2:
\[ m(-\omega^2 x) = -kx \] \[ m\omega^2 x = kx \] \[ \omega^2 = \frac{k}{m} \] \[ \omega = \sqrt{\frac{k}{m}} \]Relate Angular Frequency to Time Period: The time period \(T\) of oscillation is related to the angular frequency \(\omega\) by the formula:
\[ T = \frac{2\pi}{\omega} \]Substitute and Solve for Time Period: Substitute the expression for \(\omega\) from step 3 into the formula from step 4:
\[ T = \frac{2\pi}{\sqrt{\frac{k}{m}}} \] \[ T = 2\pi \sqrt{\frac{m}{k}} \]This is the standard formula for the time period of a spring-mass system or any system undergoing SHM with a given mass and force constant.
D) The derived formula \(T = 2\pi\sqrt{m/k}\) matches option D.
To determine where the kinetic energy is maximum in Simple Harmonic Motion (SHM), we need to understand the relationship between kinetic energy, potential energy, and velocity. In SHM, the total mechanical energy is conserved, and it continuously transforms between kinetic and potential energy. Kinetic energy is directly proportional to the square of the velocity.
Understanding SHM: In Simple Harmonic Motion, an oscillating particle moves back and forth about a fixed equilibrium position (mean position).
Velocity in SHM: The velocity \(v\) of a particle in SHM at a displacement \(x\) from the mean position is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]where \(\omega\) is the angular frequency and \(A\) is the amplitude of the oscillation.
Kinetic Energy in SHM: The kinetic energy (\(KE\)) of the particle is given by:
\[ KE = \frac{1}{2} m v^2 \]Substituting the expression for \(v\):
\[ KE = \frac{1}{2} m (\omega \sqrt{A^2 - x^2})^2 \]\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]Maximizing Kinetic Energy: For kinetic energy to be maximum, the term \((A^2 - x^2)\) must be maximum. This occurs when \(x^2\) is minimum. The minimum possible value for \(x^2\) is 0, which means \(x=0\).
Position for Maximum KE: \(x=0\) corresponds to the mean position (equilibrium position) of the oscillation. At this point, the velocity is maximum, and consequently, the kinetic energy is maximum.
Correct Option: C) Mean position
At the mean position (\(x=0\)), the velocity of the oscillating particle is maximum, and therefore, its kinetic energy is maximum. The potential energy is minimum (zero if the reference point is chosen at the mean position).