A shooting game involves using a gun that fir | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

either $1$ or $5$

Vedclass · Answer

(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.

Similar Questions

$A$ particle executing $SHM$ of amplitude $4 \, cm$ and $T = 4 \, s$. The time taken by it to move from $+2 \, cm$ to $+2\sqrt{3} \, cm$ is

$Y = A \sin (\omega t + \phi_{0})$ is the time-displacement equation of a $SHM$. At $t = 0$,the displacement of the particle is $Y = \frac{A}{2}$ and it is moving in the negative direction. Then the initial phase angle $\phi_{0}$ will be ...... .

The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

If $x = a \sin \left( \omega t + \frac{\pi}{6} \right)$ and $x' = a \cos \omega t$,then what is the phase difference between the two waves?

Vedclass Test Series

Exam Paper Generator

Online Exam Module