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 \(1.5A = \frac{3A}{2}\).