CBSE-XI-Physics
12: Simple Harmonic Motion
- #2The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitude 2 cm, 1 m s-1 and 10 m s-2 at a certain instant. Find the amplitude and the time period of the motion.Ans : It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms-1.
Acceleration of the particle, a = 10 ms-2.
Let `` \omega `` be the angular frequency of the particle.
The acceleration of the particle is given by,
a = ω2x
`` \Rightarrow \omega =\sqrt{\frac{a}{x}}=\sqrt{\frac{10}{0.02}}``
`` =\sqrt{500}=10\sqrt{5}\,\mathrm{\,Hz\,}``
`` \,\mathrm{\,Time\,}\,\mathrm{\,period\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,motion\,}\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,as\,},``
`` T=\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,\omega \,}}=\frac{2\,\mathrm{\,\pi \,}}{10\sqrt{5}}``
`` =\frac{2\times 3.14}{10\times 2.236}``
`` =0.28\,\mathrm{\,s\,}``
Now, the amplitude A is calculated as,
`` v=\omega \sqrt{{A}^{2}-{x}^{2}}``
`` ``
`` \Rightarrow {v}^{2}={\omega }^{2}\left({A}^{2}-{x}^{2}\right)``
`` 1=500\left({A}^{2}-0.0004\right)``
`` \Rightarrow A=0.0489=0.049\,\mathrm{\,m\,}``
`` \Rightarrow A=4.9\,\mathrm{\,cm\,}``
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