# Answer to Question #117966 in Mechanics | Relativity for sridhar

Question #117966
A body is in simple harmonic motion with time period T=0.5s and amplitude A=1cm. Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.
Ans 12cm/s
1
2020-05-26T12:42:37-0400

Given that, a body in simple harmonic motion with time period "T=0.5s" and and amplitude is "A=1cm=10^{-2}m" .

The equation of simple harmonic motion will be,

Let at "t=t_1" ,the position of the body from equilibrium is "\\frac{A}{2}" ,thus from above equation we get,

"\\frac{A}{2}=A\\sin(\\frac{2\\pi}{T}t_1)\\\\\n\\implies \\sin(\\frac{2\\pi}{T}t_1)=\\frac{1}{2}\\\\\n\\implies \\frac{2\\pi}{T}t_1=\\frac{\\pi}{6}\\\\\n\\implies t_1=\\frac{T}{12}\\hspace{1cm}(\\clubs)"

Clearly, if we differentiate "(\\spades)" w.r.t time we get the expression for velocity of the body,thus

"v=\\frac{2\\pi A}{T}\\cos(\\frac{2\\pi }{T}t)"

Hence,

"<v>=\\frac{1}{t_1-0}\\int_{0}^{t_1}vdt\\\\\n\\implies <v>=\\frac{1}{t_1-0}\\int_{0}^{t_1}\\frac{2\\pi A}{T}\\cos(\\frac{2\\pi }{T}t)dt\\\\\n\\implies <v>=\\frac{2\\pi A}{Tt_1}\\int_{0}^{t_1}\\cos(\\frac{2\\pi }{T}t)dt\\\\\n\\implies <v>=\\frac{2\\pi A}{Tt_1}\\frac{T}{2\\pi}\\bigg[\\sin(\\frac{2\\pi }{T}t)\\bigg]_{0}^{t_1}\\\\\n\\implies <v>=\\frac{A}{t_1}\\sin(\\frac{\\pi }{6})=\\frac{A}{2t_1}=12cm\/s"

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