Tag: angular simple harmonic motion

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Questions Related to angular simple harmonic motion

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A large box is accelerated up the inclined plane with an acceleration a and pendulum is kept vertical (Somehow by an external agent) as shown in figure.Now if the pendulum is set free to oscillate from such position, then what is the tension in the string immediately after the pendulum is set free? (mass of $500m$)

  1. $mg$
  2. $ma _{o} \sin\theta$
  3. $\left( m g + m a _ { 0 } \sin \theta \right)$

  4. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A bullet of mass $'m'$ hits a pendulum bob of mass $'2m'$ with a velocity $'v'$ and comes out of the bob with velocity $v/2$. Length of the pendulum is $2$ meter and $g=10 ms^{-2}$. The minimum value of $'v'$ for the bullet so that the bob may complete one revolution in the verticle is

  1. $40 ms^{-1}$
  2. $2.20 ms^{-1}$
  3. $3.15 ms^{-1}$
  4. $10 ms^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of momentum, the bob's velocity after impact is v/4. To complete a vertical circle, the bob must have a minimum velocity of sqrt(5gl) at the bottom, which is sqrt(5 * 10 * 2) = 10 m/s. Setting v/4 = 10 gives v = 40 m/s.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A pendulum of mass $m$ hangs from a support fixed to a trolley. The direction of the string (i.e.., angle $\theta$) when the trolley rolls up a plane of inclination $\alpha$ with acceleration $'a'$ is

  1. Zero
  2. $\tan^{-1} \alpha$
  3. $\tan^{-1}\dfrac{a+g \sin \alpha}{g \cos \alpha}$
  4. $\tan^{-1}\dfrac{a}{g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the non-inertial frame of the trolley, the bob experiences a pseudo-force 'ma' directed down the plane and gravity 'mg'. The effective acceleration vector is the sum of these components perpendicular and parallel to the plane, resulting in the angle tan(theta) = (a + g*sin(alpha)) / (g*cos(alpha)).

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The pendulum of a certain clock has time period $2.04 s$. How fast or slow does the clock run during $24$ hour?

  1. $28.8$ minutes slow
  2. $28.8$ minutes fast
  3. $14.4$ minutes fast
  4. $14.4$ minutes slow
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The clock loses time because its period is longer than the standard 2 seconds. The time lost in one day is (delta_T / T_actual) * 86400 seconds. With delta_T = 0.04s and T = 2s, the loss is (0.04/2.04) * 86400 approx 1694 seconds, which is about 28.2 minutes.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A bob is suspended from an ideal string of length $l$. Now it is pulled to a side through $60^{o}$ to vertical and rotates along a horizontal circle. Then its period of revolution is

  1. $ 2\pi \sqrt{ l/g }$
  2. $ \pi \sqrt{ l/2g }$
  3. $\pi\sqrt{ 2l/g }$
  4. $\pi\sqrt{ l/g }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a conical pendulum, the period is T = 2*pi*sqrt(h/g), where h = l*cos(theta). For theta = 60 degrees, cos(60) = 1/2, so h = l/2. Thus, T = 2*pi*sqrt(l/(2g)) = pi*sqrt(2l/g).

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Two simple pendulums have time period $4\ s$ and $5\ s$ respectively. If they started simultaneously from the mean positive in the same direction, then the phase difference between them by the time the larger one completes one osicillation is

  1. $\dfrac {\pi}{6}$
  2. $\dfrac {\pi}{3}$
  3. $\dfrac {\pi}{2}$
  4. $\dfrac {\pi}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The time periods are T1 = 4 s and T2 = 5 s. The larger one is T2 = 5 s, and in one complete oscillation of T2, time t = 5 s passes. The phase of the first pendulum in time t is omega_1 * t = (2*pi / T1) * t = (2*pi / 4) * 5 = 5*pi/2, which is 2*pi + pi/2, giving a phase of pi/2. The phase of the second pendulum is 2*pi, so the phase difference is pi/2 - 0 = pi/2.