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:
1. \(2\pi\text{ s}\)
2. \(3\pi\text{ s}\)
3. \(6\pi\text{ s}\)
4. \(9\pi\text{ s}\)
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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:
1. \(\frac{\pi}{100}\text{ s}\)
2. \(\frac{\pi}{50}\text{ s}\)
3. \(\frac{\pi}{20}\text{ s}\)
4. \(\frac{\pi}{10}\text{ s}\)
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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:
1. \(\sqrt{3}T\)
2. \(\frac{T}{\sqrt{3}}\)
3. 3T
4. \(\frac{T}{3}\)
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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:
1. \(x = \frac{A}{2\sqrt{2}}\)
2. \(x = \frac{A}{8}\)
3. \(x = \frac{A}{3}\)
4. \(x = \frac{A}{9}\)
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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, \(x = \frac{A}{3}\).
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?
1. \(60^\circ\)
2. \(72^\circ\)
3. \(90^\circ\)
4. \(120^\circ\)
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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:
1. \(\frac{7A}{3}\)
2. \(\frac{4A}{3}\)
3. \(\frac{3A}{2}\)
4. \(\frac{5A}{2}\)
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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 \(1.5A = \frac{3A}{2}\).
A particle executes SHM according to equation \(2\frac{d^2x}{dt^2} + 100x = 0\) (where \(x\) is in m and \(t\) is in second). Its time period of oscillation is
1. \(\frac{5\pi}{\sqrt{2}}\text{ s}\)
2. \(5\sqrt{2}\pi\text{ s}\)
3. \(2\sqrt{5}\pi\text{ s}\)
4. \(\frac{\sqrt{2}}{5}\pi\text{ s}\)
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Divide the given equation by 2 to get \(\frac{d^2x}{dt^2} + 50x = 0\). Comparing this with the standard SHM differential equation \(\frac{d^2x}{dt^2} + \omega^2x = 0\) gives \(\omega = \sqrt{50} = 5\sqrt{2}\text{ rad/s}\). The time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5\sqrt{2}} = \frac{\sqrt{2}}{5}\pi\text{ s}\).