The displacement of a particle executing simple harmonic motion is $y = A \sin (2t + \phi) \ m$,where $t$ is time in seconds and $\phi$ is the phase angle. At time $t = 0$,the displacement and velocity of the particle are $2 \ m$ and $4 \ ms^{-1}$ respectively. The phase angle $\phi$ is: (in $^{\circ}$)
- A$60$
- B$30$
- C$45$
- D$90$
Explore More
Similar Questions
The acceleration of a particle starting from rest varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be
Medium
View Solution$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
DifficultKCET 2011
View Solution$A$ particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude,is ............. $s$
The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.
$A$ particle executes $S.H.M.$ between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is $T_1$,and from $x = A/2$ to $x = A$ is $T_2$. Then:
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo