1
JEE Main 2021 (Online) 31st August Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
For a body executing S.H.M. :

(1) Potential energy is always equal to its K.E.

(2) Average potential and kinetic energy over any given time interval are always equal.

(3) Sum of the kinetic and potential energy at any point of time is constant.

(4) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below :
A
(3) and (4)
B
only (3)
C
(2) and (3)
D
only (2)
2
JEE Main 2021 (Online) 27th August Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.

JEE Main 2021 (Online) 27th August Morning Shift Physics - Simple Harmonic Motion Question 60 English
The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :
A
JEE Main 2021 (Online) 27th August Morning Shift Physics - Simple Harmonic Motion Question 60 English Option 1
B
JEE Main 2021 (Online) 27th August Morning Shift Physics - Simple Harmonic Motion Question 60 English Option 2
C
JEE Main 2021 (Online) 27th August Morning Shift Physics - Simple Harmonic Motion Question 60 English Option 3
D
JEE Main 2021 (Online) 27th August Morning Shift Physics - Simple Harmonic Motion Question 60 English Option 4
3
JEE Main 2021 (Online) 27th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = {T \over 4}s$$ starting from mean position. Assume that the initial phase of the oscillation is zero.
A
0.62 J
B
6.2 $$\times$$ 10$$-$$3 J
C
1.2 $$\times$$ 103 J
D
6.2 $$\times$$ 103 J
4
JEE Main 2021 (Online) 27th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$$y = {a \over 2}$$
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