(B) In simple harmonic motion $(SHM)$,the velocity of the object is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $x$ is the d

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(B) In simple harmonic motion $(SHM)$,the velocity of the object is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $x$ is the displacement from the mean position.At the extreme positions $(x = \pm A)$,the velocity $v$ is zero.Since the target spends more time in the regions where its speed is minimum (near the extreme positions),the probability of the gun hitting the target is highest in these regions.In the given figure,regions $1$ and $5$ correspond to the extreme positions of the oscillation.Therefore,the player should aim at either region $1$ or region $5$ to maximize the number of hits.

Class 11 Physics Oscillations Position of a Particle in SHM, Displacement and Phase

Medium

$A$ shooting game involves using a gun that fires by itself at random times. The player can only point the gun in a fixed direction while the target moves from side to side with simple harmonic motion,as shown. At which of the given regions should the player aim in order to score the greatest number of hits?

A
$3$
B
either $1$ or $5$
C
either $2$ or $4$
D
any of $1, 2, 3,$ or $5$

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Class 11 Questions

Class 11

Physics Questions

Chapter

Oscillations

Subtopic

Position of a Particle in SHM, Displacement and Phase

In the following table,time is in column-$I$ and the phase of an oscillator starting from the mean position is in column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ $t = \frac{T}{8}$ $(i)$ $\theta = \frac{5\pi}{4}$

$(b)$ $t = \frac{5T}{8}$ $(ii)$ $\theta = \frac{3\pi}{2}$

$(iii)$ $\theta = \frac{\pi}{4}$

Column-$I$	Column-$II$
$(a)$ $t = \frac{T}{8}$	$(i)$ $\theta = \frac{5\pi}{4}$
$(b)$ $t = \frac{5T}{8}$	$(ii)$ $\theta = \frac{3\pi}{2}$
	$(iii)$ $\theta = \frac{\pi}{4}$

Easy

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$A$ particle performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is:

MediumJEE MAIN 2021

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Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?

Difficult

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$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?

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$A$ particle oscillates in a straight line simple harmonically with a period of $8 \ s$ and an amplitude of $4 \sqrt{2} \ m$. The particle starts from the mean position. The ratio of the distance travelled by it in the $1^{\text{st}}$ second of its motion to that in the $2^{\text{nd}}$ second is: $(\sin 45^{\circ} = 1 / \sqrt{2}, \sin \frac{\pi}{2} = 1)$

MediumMHT CET 2025

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$A$ particle performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is:

Two particles are executing $S.H.M.$ of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions,each time their displacement is half of their amplitude. What is the phase difference between them?

$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?

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