CBSE-XI-Physics
12: Simple Harmonic Motion
- #1A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.Ans : It is given,
Amplitude of the simple harmonic motion, A =10 cm
At t = 0 and x = 5 cm,
Time period of the simple harmonic motion, T = 6 s
Angular frequency (ω) is given by,
`` \omega =\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,T\,}}=\frac{2\,\mathrm{\,\pi \,}}{6}=\frac{\,\mathrm{\,\pi \,}}{3}{\,\mathrm{\,sec\,}}^{-1}``
Consider the equation of motion of S.H.M,
Y = Asin `` \left(\omega t+\varphi \right)`` ...(1)
where Y is displacement of the particle, and
`` \varphi `` is phase of the particle.
On substituting the values of A, t and ω in equation (1), we get:
5 = 10sin(ω × 0 + ϕ)
`` \Rightarrow ``5 = 10sin Ï•
`` \,\mathrm{\,sin\,}\,\mathrm{\,\varphi \,}=\frac{1}{2}``
`` \Rightarrow \,\mathrm{\,\varphi \,}=\frac{\,\mathrm{\,\pi \,}}{6}``
∴ Equation of displacement can be written as,
`` x=\left(10\,\mathrm{\,cm\,}\right)\,\mathrm{\,sin\,}\left(\frac{\,\mathrm{\,\pi \,}}{3}t+\frac{\,\mathrm{\,\pi \,}}{6}\right)``
(ii) At t = 4 s,
`` x=10\,\mathrm{\,sin\,}\left[\frac{\,\mathrm{\,\pi \,}}{3}4+\frac{\,\mathrm{\,\pi \,}}{6}\right]``
`` =10\,\mathrm{\,sin\,}\left[\frac{8\,\mathrm{\,\pi \,}+\,\mathrm{\,\pi \,}}{6}\right]``
`` =10\,\mathrm{\,sin\,}\left(\frac{9\,\mathrm{\,\pi \,}}{6}\right)``
`` =10\,\mathrm{\,sin\,}\left(\frac{3\,\mathrm{\,\pi \,}}{2}\right)``
`` =10\,\mathrm{\,sin\,}\left(\pi +\frac{\,\mathrm{\,\pi \,}}{2}\right)``
`` =-10\,\mathrm{\,sin\,}\frac{\,\mathrm{\,\pi \,}}{2}=-10``
Acceleration is given by,
a = -ω2x
`` =\left(\frac{-{\,\mathrm{\,\pi \,}}^{2}}{9}\right)\times \left(-10\right)``
`` =10.9\approx 11\,\mathrm{\,cm\,}/{\,\mathrm{\,sec\,}}^{-2}``
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