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$90$
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The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in the $1^{\text{st}}$ second to the $2^{\text{nd}}$ second is:
Medium
View SolutionThe displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:
Medium
View Solution$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:
MediumIIT 2001
View Solution$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.
$A$ particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$. The angular frequency of the particle is $\pi \, rad/s$,then its amplitude is
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