Question 11:
moderateA periodic time of a body executing simple harmonic motion is 3s. After how much interval from time t = 0, its displacement from mean position will be half of its amplitude ?
\[ x= A sin\left( \omega t \right) \]
\[ \frac{A}{2}= A sin\left( \omega t \right) \]
\[ \frac{1}{2}= sin\left( \omega t \right) \]
\[ \frac{\Pi}{6}= \omega t =\frac{2\Pi}{T}t = \frac{2\Pi}{3}t \]
\[ t= \frac{1}{4}s \]
Question 12:
moderateDisplacement-time graph of a particle executing SHM is as shown below :-
The corresponding force-time graph of the particle can be :
Equation of displacement x = A sin (ωt) so, acceleration a= -Aω sin(ωt) force will have same nature so,
Question 13:
moderateA 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, \(mgL(1 - \cos\theta) = \frac{1}{2}mv^2 = mgL ⇒\cos\theta = 0 ⇒ \theta = 90^\circ\).
Question 14:
moderateEquation of SHM of a particle whose amplitude is 0.1 m and frequency is 25 Hz with an initial phase of \(\frac{\pi}{4}\) radians is
Using standard SHM formula \(x = A sin(\omega t + phi)\), where \(A = 0.1 \text{m}\), \(\omega = 2\pi f = 2\pi(25) = 50\pi \text{rad/s}\), and \(\phi = \frac{\pi}{4}\). Substituting gives \(x = 0.1 sin \left(50\pi t + \frac{\pi}{4}\right)\).
Question 15:
moderateA body is vibrating with SHM of amplitude 15 cm and frequency 4 Hz. The maximum value of acceleration is
The maximum acceleration in SHM is given by \(a_{\text{max}} = \omega^2 A = (2\pi f)^2 A\). Substituting \(f = 4\text{ Hz}\) and \(A = 0.15\text{ m}\), we find \(a_{\text{max}} = (8\pi)^2 \times 0.15 \approx 94.65 \text{ m s}^{-2}\).
Question 16:
moderateThe displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
The expression represents two perpendicular SHMs of the same frequency with a phase difference of \(\frac{\pi}{2}\). The resultant amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).
Question 17:
moderateA block is resting on a piston which is moving vertically executing SHM of period 1 s. At what minimum amplitude of motion, will the block and piston separate? (take \(\pi^2 = 10\))
Separation occurs when the maximum downward acceleration of the piston equals \(g\). Thus, \(\omega^2 A = g \implies \left(\frac{2\pi}{T}\right)^2 A = g \implies 4\pi^2 A = 10 \implies 40 A = 10 \implies A = 0.25\text{ m}\).
Question 18:
moderateIf \( x = 2\sin\left(\frac{\pi}{2}t\right) \) represents the motion of a particle executing SHM, the maximum speed of the particle in \( \text{m s}^{-1} \) is (All parameters are in SI units)
Comparing the given equation with the standard SHM equation \( x = A\sin(\omega t) \), we get \( A = 2\text{ m} \) and \( \omega = \frac{\pi}{2}\text{ rad/s} \). The maximum speed is \( v_{\max} = A\omega = 2 \times \frac{\pi}{2} = \pi\text{ m/s} \).
Question 19:
moderateA particle is executing S.H.M. along a straight line. Its velocities at distances $ x_1 $ and $ x_2 $ from the mean position are $ v_1 $ and $ v_2 $, respectively. Its time period is:
(2015)
Velocity in SHM is $ v^2 = \omega^2(A^2 - x^2) $. So, $ v_1^2 = \omega^2(A^2 - x_1^2) $ and $ v_2^2 = \omega^2(A^2 - x_2^2) $. Subtracting these equations gives $ v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2) $, yielding $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $$.
Question 20:
moderateThe displacement of a particle along the x-axis is given by $x = a\sin^2\omega t$. The motion of the particle corresponds to:
(2010 Pre)
Equation is $x = a\sin^2\omega t = \frac{a}{2}(1 - \cos 2\omega t)$. This represents SHM about the mean position $x = a/2$. The angular frequency is $2\omega$. The frequency is $f = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.