the time period of oscillations of a simple pendulum depends on the length of the pendulum mass of the bob and the acceleration due to gravity.deduce the relation among them by the method of dimensions
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hey mate !! here is the answer...
T has the dimension [T]
now according to the law of dimensions the dimension of LHS=RHS.
[T] is directly proportional to [L] ^a[LT^-2]^b
[T] is directly proportional to [L]^a+b[T]^-2b
by comparison of LHS and RHS we get
1=-2b
=b= -1/2
and
a+b= 0
a-1/2=0
a=1/2
from the above datum we get T= k(l/g)^1/2
hoping for fovourable response..
thanking you.
hope it helps you .
T has the dimension [T]
now according to the law of dimensions the dimension of LHS=RHS.
[T] is directly proportional to [L] ^a[LT^-2]^b
[T] is directly proportional to [L]^a+b[T]^-2b
by comparison of LHS and RHS we get
1=-2b
=b= -1/2
and
a+b= 0
a-1/2=0
a=1/2
from the above datum we get T= k(l/g)^1/2
hoping for fovourable response..
thanking you.
hope it helps you .
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