1
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
Out of Syllabus
The displacement of a damped harmonic oscillator is given by
x(t ) = e–0.1t cos (10$$\pi$$t + f).
Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :
A
27 s
B
13 s
C
7 s
D
4 s
2
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
Out of Syllabus
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
3
JEE Main 2019 (Online) 12th January Evening Slot
+4
-1
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
4
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
$${1 \over {2\pi }}\sqrt {{{2k} \over m}}$$
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