The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is :
\[ x= A sin\left( \omega t \right) \]
x= (1cm) sin (2π/8t)= (1 cm ) sin ( π/4t)
v=dx/dt= π/4 cos( π/4t)
a= dv/dt = -(π/4)²sin (π/4t)
a= -(π/4)²sin (π/4×4/3)= -√3/32π² cm/s²
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 \(T = \frac{2\pi}{\omega} = 6\pi\text{ s}\).
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.
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 \(1.5A = \frac{3A}{2}\).
A simple pendulum oscillates in a vertical plane. When it passes through the mean position, the tension in the string is \(3\) times the weight of the pendulum bob. What is the maximum angular displacement of the pendulum of the string with respect to the vertical ?
At the mean position, tension is \(T = mg + \frac{mv^2}{L}\). Given \(T = 3mg ⇒ \frac{mv^2}{L} = 2mg ⇒ v^2 = 2gL\). Using conservation of energy, \(mgL(1 - \cos\theta) = \frac{1}{2}mv^2 = mgL ⇒\cos\theta = 0 ⇒ \theta = 90^\circ\).
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
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 \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5\sqrt{2}} = \frac{\sqrt{2}}{5}\pi\text{ s}\).
A body oscillates in SHM according to the equation \( x = 10 cos \left(2\pi t + \frac{\pi}{4} \right)\text{ cm}\). Its instantaneous displacement at \(t = 1\text{ s}\) is
Substituting \(t = 1\text{ s}\) in the given equation: \[x = 10 cos\left(2\pi(1) + \frac{\pi}{4}\right) = 10 cos\left(\frac{\pi}{4}
/right) = \frac{10}{\sqrt{2}} = 5\sqrt{2}\text{ cm}\].
Which of the following examples does not represent SHM?
Rotation of the Earth about its axis is a periodic motion but not oscillatory. Since it has no restoring force or back-and-forth motion about a mean position, it is not simple harmonic motion.
The distance covered by a particle undergoing SHM in one time period is (A = amplitude of oscillation)
In one complete time period, the particle travels from mean position to one extreme, back to mean, to the other extreme, and back to mean. Total distance is \(A + A + A + A = 4A\).