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:
1. \(\frac{7A}{3}\)
2. \(\frac{4A}{3}\)
3. \(\frac{3A}{2}\)
4. \(\frac{5A}{2}\)
View Answer
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 \(1.5A = \frac{3A}{2}\).
A 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
1. \(\frac{5\pi}{\sqrt{2}}\text{ s}\)
2. \(5\sqrt{2}\pi\text{ s}\)
3. \(2\sqrt{5}\pi\text{ s}\)
4. \(\frac{\sqrt{2}}{5}\pi\text{ s}\)
View Answer
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 \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5\sqrt{2}} = \frac{\sqrt{2}}{5}\pi\text{ s}\).
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
1. 10 cm
2. \(\frac{5}{\sqrt{2}}\text{ cm}\)
3. \[5\sqrt{2}\text{ cm}\]
4. \[10\sqrt{2}\text{ cm}\]
View Answer
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?
1. Oscillations of a spring block system
2. Motion of ball bearing inside smooth curved bowl, when released slightly away from equilibrium position
3. Motion of oscillating mercury column in vertical U-tube
4. Rotation of earth about its own axis
View Answer
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.
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
1. 5 s
2. 2 s
3. \( \frac{1}{5} \, \text{s} \)
4. \( \frac{1}{10} \, \text{s} \)
View Answer
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
1. \( 4\text{ rad/s} \)
2. \( 3\text{ rad/s} \)
3. \( 2\text{ rad/s} \)
4. \( 4pi\text{ rad/s} \)
View Answer
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.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.