A particle is oscillating according to the eq | vedclass

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

Question

(C) The given equation of motion is $X = 7 \cos(0.5 \pi t)$.Comparing this with the standard $SHM$ equation $X = A \cos(\omega t)$,we get the angular

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The given equation of motion is $X = 7 \cos(0.5 \pi t)$.Comparing this with the standard $SHM$ equation $X = A \cos(\omega t)$,we get the angular frequency $\omega = 0.5 \pi \ rad/s$.The time period $T$ is given by $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{0.5 \pi} = 4 \ s$.The time taken by a particle to move from the equilibrium position (mean position) to the maximum displacement (extreme position) is one-fourth of the time period.Therefore,$t = \frac{T}{4} = \frac{4}{4} = 1 \ s$.

$A$ particle is oscillating according to the equation $X = 7 \cos(0.5 \pi t)$,where $t$ is in seconds. The particle moves from the position of equilibrium to maximum displacement in time ..... $s$.

Similar Questions

What is the ratio between the distance travelled by the oscillator in one time period and the amplitude?

$Assertion :$ In simple harmonic motion,the motion is to and fro and periodic.
$Reason :$ Velocity of the particle $(v) = \omega \sqrt {A^2 - x^2}$ (where $x$ is the displacement and $A$ is the amplitude).

Define periodic motion and oscillatory motion.

Who decides the characteristics of $SHM$?

The displacement of a particle varies according to the relation $x = 3 \sin 100t + 8 \cos^2 50t$. Which of the following is incorrect about this motion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module