1
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-
A
100 s
B
10 s
C
20 s
D
50 s
2
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A simple harmonic motion is represented by :

y = 5 (sin 3 $$\pi $$ t + $$\sqrt 3 $$ cos 3 $$\pi $$t) cm

The amplitude and time period of the motion are :
A
10 cm, $${3 \over 2}$$ s
B
5 cm, $${2 \over 3}$$ s
C
5 cm, $${3 \over 2}$$ s
D
10 cm, $${2 \over 3}$$ s
3
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Simple Harmonic Motion Question 101 English
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
$${1 \over {2\pi }}\sqrt {{{2k} \over m}} $$
4
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
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
1 rad/s
B
10$$-$$3 rad/s
C
10$$-$$1 rad/s
D
10$$-$$5 rad/s
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