1
JEE Main 2021 (Online) 25th February Evening Shift
+4
-1 Y = A sin($$\omega$$t + $$\phi$$0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $$Y = {A \over 2}$$ and it is moving along negative x-direction. Then the initial phase angle $$\phi$$0 will be:
A
$${{5\pi } \over 6}$$
B
$${{\pi } \over 3}$$
C
$${{2\pi } \over 3}$$
D
$${{\pi } \over 6}$$
2
JEE Main 2021 (Online) 25th February Evening Shift
+4
-1 Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is : A
$$2\pi \sqrt {{m \over k}}$$
B
$$\pi \sqrt {{m \over k}}$$
C
$$2\pi \sqrt {{m \over {2k}}}$$
D
$$\pi \sqrt {{m \over {2k}}}$$
3
JEE Main 2021 (Online) 25th February Evening Shift
+4
-1 The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30$$^\circ$$ in 0.1 s. The perpendicular projection 'P' from 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be : A
9.87 N
B
0.49 N
C
50 N
D
100 N
4
JEE Main 2021 (Online) 25th February Morning Slot
+4
-1
If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
A
16 m/s2
B
2$$\pi$$2 ms$$-$$2
C
$$\pi$$2 ms$$-$$2
D
9.8 ms$$-$$2
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