What should be the displacement of a simple pendulum whose amplitude is A, at which potential energy is 1/4 th of the total energy ?
\[ \frac{1}{2} k x^{2}=\frac{1}{4}\left( \frac{1}{2}kA^{2} \right) \]
\[ \frac{1}{2}k x^{2}=\frac{1}{4}\left(kA^{2} \right) x= \frac{A}{2} \]
What is the angular frequency of the system shown in the figure?
The system shown consists of two masses \( M \) connected by a spring with a spring constant \( k \). Since the masses are identical, the angular frequency \( \omega \) of the system for oscillations is given by:
\[
\omega = \sqrt{\frac{k}{\text{reduced mass}}}
\]
In this case, the reduced mass \( \mu \) of the system is given by:
\[
\mu = \frac{M \cdot M}{M + M} = \frac{M}{2}
\]
Thus, the angular frequency \( \omega \) is:
\[
\omega = \sqrt{\frac{k}{M/2}} = \sqrt{\frac{2k}{M}}
\]
Answer:
\[
\omega = \sqrt{\frac{2k}{M}}
\]
For a particle executing simple harmonic motion, the amplitude is \(A\) and time period is \(T\). The maximum speed will be:
The maximum speed of a particle in simple harmonic motion is given by \(v_{\text{max}} = A\omega\). Since \(\omega = \frac{2\pi}{T}\), we get \(v_{\text{max}} = \frac{2\pi A}{T}\).
A particle is performing SHM along x-axis such that its velocity and displacement are related as \(27v^2 = 10 – 3x^2\), then time period of oscillation of particle is:
The given equation can be rewritten as \(v^2 = \frac{10}{27} - \frac{1}{9}x^2\). Comparing this with the standard SHM equation \(v^2 = \omega^2(A^2 - x^2)\), we get \(\omega^2 = \frac{1}{9}\) which gives \(omega = \frac{1}{3}\text{ rad/s}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = 6\pi\text{ s}\).
The potential energy of a harmonic oscillator of mass \(2\text{ kg}\) in its mean position is \(5\text{ J}\). If its total energy is \(9\text{ J}\) and its amplitude is \(0.01\text{ m}\), its time period will be:
The total energy is given by \(E = U_0 + \frac{1}{2}m\omega^2 A^2\). Substituting the values: \(9 = 5 + \frac{1}{2}(2)\omega^2 (0.01)^2\), which gives \(\omega = 200\text{ rad/s}\). Therefore, the time period is \(T = \frac{2\pi}{\omega} = \frac{\pi}{100}\text{ s}\).
When a block is suspended from a spring, time period of its oscillation is \(T\). If this spring is cut into 3 equal parts and this block is suspended from parallel combination of these 3 parts, then new time period of oscillation will be:
Cutting a spring of constant \(k\) into 3 equal parts increases each part's spring constant to \(3k\). In a parallel connection, the equivalent constant is \(k_{\text{eq}} = 3k + 3k + 3k = 9k\). The new time period is \(T' = 2\pi\sqrt{\frac{m}{9k}} = \frac{T}{3}\).
A particle is performing simple harmonic motion of amplitude \(A\) about origin. Then position at which kinetic energy of particle is 8 times of its potential energy at that instant is:
The given condition is \(\text{KE} = 8\text{PE}\). Substituting the expressions: \(\frac{1}{2}m\omega^2(A^2 - x^2) = 8\left(\frac{1}{2}m\omega^2 x^2\right)\), which simplifies to \(A^2 - x^2 = 8x^2\) or \(9x^2 = A^2\). Thus, \(x = \frac{A}{3}\).
A particle is executing SHM. Then, the graph of velocity as a function of displacement is a/an:
For a particle in SHM, velocity \(v = \omega \sqrt{A^2 - x^2}\). Squaring and rearranging gives \(\frac{v^2}{\omega^2 A^2} + \frac{x^2}{A^2} = 1\), which represents an ellipse.
Two pendulums have time periods T and \(\frac{5T}{4}\). They start SHM at the same time from the mean position. What will be the phase difference between them, when the smaller pendulum has completed one oscillation?
For the smaller pendulum, time elapsed for one oscillation is \(t = T\), so its phase is \(2\pi\). The phase of the second pendulum is \(\phi_2 = \frac{2\pi}{5T/4}T = \frac{8\pi}{5}\). The phase difference is \(2\pi - \frac{8\pi}{5} = \frac{2\pi}{5} = 72^\circ\).
A particle is performing simple harmonic motion of amplitude A and time period 12 seconds. If at t = 0 particle is at mean position, then distance travelled by particle in first 5 seconds will be:
The equation of motion is \(x = A \sin\left(\frac{\pi t}{6}\right)\). At \(t = 3\text{ s}\), \(x = A\). At \(t = 5\text{ s}\), \(x = A \sin(5\pi/6) = 0.5A\). Total distance is \(A\) (from 0 to \(A\)) plus \(0.5A\) (returning from \(A\) to \(0.5A\)), which is \(1.5A = \frac{3A}{2}\).