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JEE Main 2021 (Online) 27th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $${{3E} \over 4}$$ then its displacement 'y' is given by :
A
y = a
B
$$y = {a \over {\sqrt 2 }}$$
C
$$y = {{a\sqrt 3 } \over 2}$$
D
$$y = {a \over 2}$$
2
JEE Main 2021 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.
A
$${1 \over 2}$$
B
$${3 \over 4}$$
C
$${1 \over 3}$$
D
$${1 \over 4}$$
3
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
T0 is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $${1 \over {16}}$$ times of its initial value, the modified time period is :
A
4 T0
B
$${1 \over {4}}$$ T0
C
T0
D
8$$\pi$$ T0
4
JEE Main 2021 (Online) 20th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as :
A
$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}} $$
B
$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}} $$
C
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}} $$
D
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$
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