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Question: A particle moving with simple harmonic motion has speed 3cm/s and 4cm/s at displacement 8cm and 6c...

A particle moving with simple harmonic motion has speed 3cm/s and 4cm/s at
displacement 8cm and 6cm, respectively, from the equilibrium position. Find
(a)the period of oscillation, and
(b)the amplitude of oscillation.

Explanation

Solution

For a body performing SHM velocity at a given displacement is given by the equationv=ωA2x2v = \omega \sqrt {{A^2} - {x^2}} . We can divide the two equations for the two given displacements to get the desired variable (here amplitude A). The time period for a body performing SHM is given byT=2πωT = \dfrac{{2\pi }}{\omega }.

Step by step answer: The velocity of a body performing simple harmonic motion at a given displacement is given by the equation v=ωA2x2v = \omega \sqrt {{A^2} - {x^2}} .
Given, at a displacement x=x = 8 cm the velocity of the body performing SHM v=3cm/s = 3cm/s.
3=ωA282\therefore 3 = \omega \sqrt {{A^2} - {8^2}} ----------(1)
And for the displacement x=x = 6cm, the velocity of the body v=4cm/s = 4cm/s.
4=ωA262\therefore 4 = \omega \sqrt {{A^2} - {6^2}} ----------(2)
Dividing both equations we get,
34=ωA264ωA236 (34)2=A264A236 9A2324=16A21024 7A2=700 A2=7007=100 A=100=10  \dfrac{3}{4} = \dfrac{{\omega \sqrt {{A^2} - 64} }}{{\omega \sqrt {{A^2} - 36} }} \\\ \Rightarrow {\left( {\dfrac{3}{4}} \right)^2} = \dfrac{{{A^2} - 64}}{{{A^2} - 36}} \\\ \Rightarrow 9{A^2} - 324 = 16{A^2} - 1024 \\\ \Rightarrow 7{A^2} = 700 \\\ \Rightarrow {A^2} = \dfrac{{700}}{7} = 100 \\\ \therefore A = \sqrt {100} = 10 \\\
Therefore the amplitude of the body is 10cm.
Substituting the value of amplitude 'A' in equation 1 we get,
3=ω10282 3=ω10064 3=ω36 ω=36=12  3 = \omega \sqrt {{{10}^2} - {8^2}} \\\ \Rightarrow 3 = \omega \sqrt {100 - 64} \\\ \Rightarrow3 = \omega \sqrt {36} \\\ \therefore \omega = \dfrac{3}{6} = \dfrac{1}{2} \\\
The time period of body performing simple harmonic motion is given by T=2πωT = \dfrac{{2\pi }}{\omega }
Substituting the value of ω\omega in the equation we get T=2π12=4πT = \dfrac{{2\pi }}{{\dfrac{1}{2}}} = 4\pi
Therefore the time period =4π= 4\pi and the amplitude of the body is 10cm.

Note: For a particle performing simple harmonic motion its acceleration is directly proportional to its displacement from its mean position and is directed always opposite to its displacement. We can derive the equation of velocity from the general equation of simple harmonic motion by differentiating it.