Question 111:
easyA particle moves according to equation, \(x = a \cos \frac{\pi t}{2}\) . The distance covered by it in the time interval between t = 0 to t = 3 s is
The time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/2} = 4\text{ s}\). In \(t = 3\text{ s}\) (which is \(\frac{3T}{4}\)), the particle completes three quarters of an oscillation, covering a total distance of \(3a\).
Question 112:
easyThe displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
Since the two perpendicular components have a phase difference of \(\frac{\pi}{2}\), the net amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).
Question 113:
easyA body is vibrating with SHM of amplitude \(15\text{ cm}\) and frequency \(4\text{ Hz}\). The maximum value of acceleration is
The maximum acceleration is given by \(a_{\text{max}} = \omega^2 A = (2\pi f)^2 A\). Substituting \(f = 4\text{ Hz}\) and \(A = 0.15\text{ m}\) gives \(a_{\text{max}} = 4\pi^2 (16)(0.15) \approx 94.65\text{ m/s}^2\).
Question 114:
easyThe differential equation of motion of a particle executing SHM is \(\frac{d^2y}{dt^2} + Ky = 0\) where \(K\) is positive constant. The time period of the oscillation is given by
Comparing with the standard equation \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we find \(\omega = \sqrt{K}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).
Question 115:
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 116:
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 117:
easyA simple pendulum hanging freely stayed at rest in vertical, because in this position
A stable equilibrium state corresponds to a local minimum of the system's potential energy. For a simple pendulum, the lowest point is the vertical position, where potential energy is minimum.
Question 118:
moderateFor the damped oscillator, if time taken for its amplitude of vibrations to drop to half of its initial value is \(T\) then time taken for amplitude to drop to one eighth amplitude is
Amplitude decays exponentially as \(A = A_0 e^{-\gamma t}\). Since it halves in time \(T\), to drop to \(1/8 = (1/2)^3\) of its initial value, it takes exactly \(3T\).
Question 119:
moderateA particle executes linear simple harmonic motion with an amplitude of $ 3 \text{ cm} $. When the particle is at $ 2 \text{ cm} $ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is: (2017-Delhi)
Given $ |v| = |a| $, we have $ \omega \sqrt{A^2 - x^2} = \omega^2 x $. Substituting $ A = 3 $ and $ x = 2 $, we get $ \sqrt{3^2 - 2^2} = \omega (2) \implies \omega = \frac{\sqrt{5}}{2} $. The time period is $ T = \frac{2\pi}{\omega} = \frac{4\pi}{\sqrt{5}} $.
Question 120:
easyWhen two displacements represented by $ y_1 = a \sin(\omega t) $ and $ y_2 = b \cos(\omega t) $ are superimposed, the motion is:
(2015)
The resultant displacement is $ y = y_1 + y_2 = a \sin(\omega t) + b \cos(\omega t) $. This equation represents a single simple harmonic motion with a resultant amplitude of $ R = \sqrt{a^2 + b^2} $.