1
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length $$\ell$$ and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is :

A
$${1 \over {2\pi }}\sqrt {{{3k} \over m}}$$
B
$${1 \over {2\pi }}\sqrt {{{6k} \over m}}$$
C
$${1 \over {2\pi }}\sqrt {{k \over m}}$$
D
$${1 \over {2\pi }}\sqrt {{{2k} \over m}}$$
2
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
Out of Syllabus
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :
A
B
10$$-$$3 rad/s
C
10$$-$$1 rad/s
D
10$$-$$5 rad/s
3
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then :
A
$${K_2}$$ = $${{{K_1}} \over 2}$$
B
K2 = 2K1
C
K2 = K1
D
K2 = $${{{K_1}} \over 4}$$
4
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :
A
$${{\sqrt 3 } \over 2}$$ s
B
$${3 \over 2}$$ s
C
$${2 \over {\sqrt 3 }}$$ s
D
$$2\sqrt 3$$ s
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