1
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A spring mass system (mass m, spring constant k and natural length $$l$$) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity $$\omega $$, (k $$ \gg m{\omega ^2}$$) the relative change in the length of the spring is best given by the option :
A
$${{m{\omega ^2}} \over {3k}}$$
B
$${{m{\omega ^2}} \over k}$$
C
$${{2m{\omega ^2}} \over k}$$
D
$$\sqrt {{2 \over 3}} \left( {{{m{\omega ^2}} \over k}} \right)$$
2
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A simple pendulum is being used to determine th value of gravitational acceleration g at a certain place. Th length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :
A
4.40%
B
3.40%
C
2.40%
D
5.40%
3
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
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
4
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
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