1
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
2
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 = $${{{K_1}} \over 4}$$
3
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
4
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin$${{\pi t} \over {90}}$$. The ratio of kinetic to potential energy of this particle at t = 210 s will be:
A
$${1 \over 9}$$
B
3
C
2
D
1
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