NEET UG PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A conical pendulum of length 1 m makes an angle

\theta = {45^0}

w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m. and its centre is vertically below O. The speed of the pendulum, in its circular path, will be

\left( {g = 10\mathrm{ }m{s^{ - 2}}} \right)

View answer and explanation

Correct option: D

2 m/s

Option D (2 m/s) is correct because it's calculated using the radius of the circular path and the angle with respect to the vertical axis.

More Questions from OSCILLATIONS

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A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is

{f_1}

. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is

{f_2}

. Which of the following relations between the frequencies is correct The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be. Under the action of a force

F=-k x^3

, the motion of a particle is

(k=

a positive constant

)

S-I: A particle executes SHM with frequency

f

. The frequency with which total mechanical energy of particle oscillates is

2 f

.
S-II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency

2 f.

The phase difference (in radian) between two SHM whose equations are

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

and

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

t=0.5 \mathrm{~s}

is: A vertical spring block system is made to oscillate.
Statement-1: Its time period on earth is more than that on the moon.
Statement-2: Its extension on earth (in equilibrium) is more than that on the moon.