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 \(A + A + A + A = 4A\).
Assertion (A): In a SHM, kinetic and potential energies become equal when the displacement is \(\frac{1}{\sqrt{2}}\) times the amplitude.
Reason (R): In SHM, kinetic energy is zero when potential energy is maximum.
Equating \(KE = \frac{1}{2}k(A^2 - x^2)\) and \(PE = \frac{1}{2}kx^2\) gives \(x = \frac{A}{\sqrt{2}}\). Maximum PE occurs at extremes where KE is zero. Both statements are true, but (R) does not explain (A).
A 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 \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{4}{\pi^2}} = 2\pi \left(\frac{2}{\pi}\right) = 4 \text{ s}\).
The 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, \(\omega^2 = 4 ⇒ \omega = 2\text{ rad/s}\).
If 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 \( T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi} = \frac{1}{5} \, \text{s} \).
The 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, \( \omega^2 = 4 ⇒ \omega = 2\text{ rad/s} \).
Assertion (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.
A vertical spring block system is made to oscillate.
Assertion (A): Its time period on earth is more than that on the moon.
Reason (R): Its extension on moon (in equilibrium) is more than that on the earth.
The time period of a vertical spring-block system is \(T = 2\pi \sqrt{\frac{m}{k}}\), which is independent of gravity (g). So (A) is false. The equilibrium extension is \(x_{eq} = \frac{mg}{k}\). Since \(g_{moon} < g_{earth}\), then \(x_{eq,moon} < x_{eq,earth}\). So (R) is also false.