Question 11:
easyA 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 \(1.5A = \frac{3A}{2}\).
Question 12:
easyA 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 \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5\sqrt{2}} = \frac{\sqrt{2}}{5}\pi\text{ s}\).
Question 13:
easyA 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: \[x = 10 cos\left(2\pi(1) + \frac{\pi}{4}\right) = 10 cos\left(\frac{\pi}{4}
/right) = \frac{10}{\sqrt{2}} = 5\sqrt{2}\text{ cm}\].
Question 14:
easyWhich 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.
Question 15:
easyThe 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 \(A + A + A + A = 4A\).
Question 16:
easyA particle of mass 4 kg is moving along x-axis under the action of force \(F = -\pi^2 x\) (where F is in newton and x is in m). The time period of oscillation is
Comparing \(F = -\pi^2 x\) with \(F = -k x\), we get \(k = \pi^2\). The time period is \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{4}{\pi^2}} = 2\pi \left(\frac{2}{\pi}\right) = 4 \text{ s}\).
Question 17:
easyThe differential equation for a particle executing S.H.M. is given by \(\frac{d^2y}{dt^2} + 4y = 0\), where symbols have their usual meaning. The angular velocity of the particle is given by
The standard equation of S.H.M. is \(\frac{d^2y}{dt^2} + \omega^2 y = 0\). Comparing this with the given equation, \(\omega^2 = 4 ⇒ \omega = 2\text{ rad/s}\).
Question 18:
easyIf maximum speed and maximum acceleration of a particle executing SHM is found to be 5 cm/s and \( 50\pi \, \text{cm/s}^2 \) respectively, then its time period will be
Using \( v_{\text{max}} = A \omega = 5 \) and \( a_{\text{max}} = A \omega^2 = 50\pi \), we get \( \omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{50\pi}{5} = 10\pi \, \text{rad/s} \). The time period is \( T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi} = \frac{1}{5} \, \text{s} \).
Question 19:
easyThe differential equation for a particle executing S.H.M. is given by \( \frac{d^2 y}{dt^2} + 4y = 0 \), where symbols have their usual meaning. The angular velocity of the particle is given by
The standard differential equation of S.H.M. is \( \frac{d^2 y}{dt^2} + \omega^2 y = 0 \). By comparison, \( \omega^2 = 4 ⇒ \omega = 2\text{ rad/s} \).
Question 20:
easyAssertion (A): A hole were drilled through the centre of earth and a ball is dropped into the hole at one end, it will not get out of other end of the hole.
Reason (R): Ball will execute simple harmonic motion inside the hole.
Concept of gravity inside earth and SHM. Inside Earth, \(F = -k r\). A ball dropped in a hole through the Earth's center oscillates in SHM. It reaches the other end with zero velocity and turns back, thus not 'getting out' (escaping) the hole.
Both A and R are true, and R explains A.