1
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

The time taken by a particle executing simple harmonic motion of period '$$\mathrm{T}$$', to move from the mean position to half the maximum displacement is

A
$$\frac{\mathrm{T}}{12} \mathrm{~s}$$
B
$$\frac{\mathrm{T}}{2} \mathrm{~s}$$
C
$$\frac{\mathrm{T}}{4} \mathrm{~s}$$
D
$$\frac{\mathrm{T}}{6} \mathrm{~s}$$
2
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a medium, the phase difference between two particles separated by a distance '$$x$$' is $$\left(\frac{\pi}{5}\right)^{\text {c }}$$. If the frequency of the oscillation of particles is $$25 \mathrm{~Hz}$$ and the velocity of propagation of the waves is $$75 \mathrm{~m} / \mathrm{s}$$, then the value of $$x$$ is

A
0.4 m
B
0.1 m
C
0.2 m
D
0.3 m
3
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+1
-0

A particle starts oscillating simple harmonically from its mean position with time period '$$T$$'. At time $$t=\frac{T}{12}$$, the ratio of the potential energy to kinetic energy of the particle is $$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$$

A
$$1: 2$$
B
$$3: 1$$
C
$$2: 1$$
D
$$1: 3$$
4
MHT CET 2021 24th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

The displacement of a particle performing S.H.M. is given by $$x=5 \sin (3 t+3)$$, where $$x$$ is in $$\mathrm{cm}$$ and $$t$$ is in second. The maximum acceleration of the particle will be

A
$$15 \mathrm{~cm} \mathrm{~s}^{-2}$$
B
$$30 \mathrm{~cm} \mathrm{~s}^{-2}$$
C
$$45 \mathrm{~cm} \mathrm{~s}^{-2}$$
D
$$90 \mathrm{~cm} \mathrm{~s}^{-2}$$
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