1
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :

A
Circular
B
Elliptical
C
Sinusoidal
D
Straight line
2
JEE Main 2022 (Online) 25th July Morning Shift
MCQ (Single Correct Answer)
+4
-1

In figure $$(\mathrm{A})$$, mass '$$2 \mathrm{~m}^{\text {' }}$$ is fixed on mass '$$\mathrm{m}$$ ' which is attached to two springs of spring constant $$\mathrm{k}$$. In figure (B), mass '$$\mathrm{m}$$' is attached to two springs of spring constant '$$\mathrm{k}$$' and '$$2 \mathrm{k}^{\prime}$$. If mass '$$\mathrm{m}$$' in (A) and in (B) are displaced by distance '$$x^{\prime}$$ horizontally and then released, then time period $$\mathrm{T}_{1}$$ and $$\mathrm{T}_{2}$$ corresponding to $$(\mathrm{A})$$ and (B) respectively follow the relation.

A
$$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{\sqrt{2}}$$
B
$$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{3}{2}}$$
C
$$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{2}{3}}$$
D
$$\frac{T_{1}}{T_{2}}=\frac{\sqrt{2}}{3}$$
3
JEE Main 2022 (Online) 29th June Evening Shift
MCQ (Single Correct Answer)
+4
-1

The motion of a simple pendulum executing S.H.M. is represented by the following equation.

$$y = A\sin (\pi t + \phi )$$, where time is measured in second. The length of pendulum is

A
97.23 cm
B
25.3 cm
C
99.4 cm
D
406.1 cm
4
JEE Main 2022 (Online) 28th June Morning Shift
MCQ (Single Correct Answer)
+4
-1

Motion of a particle in x-y plane is described by a set of following equations $$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m$$ and $$y = 4\sin (\omega t)\,m$$. The path of the particle will be :

A
circular
B
helical
C
parabolic
D
elliptical
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