Derive an expression for the time period (T)of a simple pendulum which may depend upon the mass (m) of the bob , length (l) of the pendulum and acceleration due to gravity(g).

Derive an expression for the time period (T)of a simple pendulum which

1.	Derive an expression for the time period (T)of a simple pendulum which may depend upon the mass (m) of the bob , length (l) of the pendulum and acceleration due to gravity(g).
Answer» <html><body><p></p>Solution :Let `T=km^(x) l^(y)g^(<a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a>) ` where k is a <a href="https://interviewquestions.tuteehub.com/tag/dimensionless-954191" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONLESS">DIMENSIONLESS</a> constant.<br/> <a href="https://interviewquestions.tuteehub.com/tag/writing-25453" style="font-weight:bold;" target="_blank" title="Click to know more about WRITING">WRITING</a> the equation in dimensional form, we have `[M^(0)L^(0)t^(1)]=[M]^(x)[L]^(y)[LT^(-2)]^(z)=[M^(x)L^(Y+Z)T^(-<a href="https://interviewquestions.tuteehub.com/tag/2z-301338" style="font-weight:bold;" target="_blank" title="Click to know more about 2Z">2Z</a>)]`<br/> Equating exponent of M,L and T on both sides, <br/>we get `x=0,y+z=0,-2z=1`. <br/>Solving the eq., we get `x=0,y=1/2,z=-1/2` <br/> Hence , `T=ksqrt(l/g)`, where k is constant.</body></html>