Question
Assertion In SHM, the motion is to end fro and periodic. Reason Velocity of the particle $(v)=\omega \sqrt{K^{2}-x^{2}}$, where $x$ is the displacement and $k$ is amplitude.
Step 1
This is true because in SHM, a particle oscillates back and forth about a fixed point in a regular time interval. This means the motion is periodic and the particle moves to and fro. Show more…
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A particle undergoes SHM with angular frequency $\omega .$ The initial displacement is $x_{0}$ and the initial velocity is $v_{0} .$ Deduce that an expression for the amplitude of this motion is $$A=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}$$.
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$x=A \sin \omega t$, represents the equation of a SHM. If displacements of the particle are $x_{1}$ and $x_{2}$ and velocities are $v_{1}$ and $v_{2}$, respectively, then amplitude of SHM is (A) $\left[\frac{\left(v_{2}-v_{1}\right)\left(x_{2}-x_{1}\right)}{v_{2}^{2}-v_{1}^{2}}\right]^{\frac{1}{2}}$ (B) $\left[\frac{\left(v_{2} x_{1}\right)^{2}-\left(v_{1} x_{2}\right)^{2}}{v_{2}^{2}-v_{1}^{2}}\right]^{\frac{1}{2}}$ (C) $\frac{v_{1} x_{2}}{\left(v_{2}-v_{1}\right)^{2}}$ (D) $\left[\frac{v_{1} x_{2}-v_{2} x_{1}}{v_{1}^{2}-v_{2}^{2}}\right]^{\frac{1}{2}}$
Assertion The amplitude of a particle executing SHM with a frequency of $60 \mathrm{~Hz}$ is $0.01 \mathrm{~m}$. The maximum value of acceleration of the particle is $\pm 144 \pi^{2} \mathrm{~ms}^{-2}$. Reason Acceleration amplitude $=\omega^{2} A$,where $A$ is displacement amplitude.
Oscillations
Round 2
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