Time period is the time taken by a complete oscillation or a complete cycle of a periodic motion. It is generally measured in seconds (s). In this question, we are supposed to find out the dimensions of time period using density (D), acceleration (a) and force (F) as basic quantities.


To find the dimensions of time period using D, a and F, we can use the method of dimensional analysis.


The formula of time period (T) is given by:
T = 2π√(m/k)
where m is the mass of the object in motion and k is the spring constant of the spring.


We can substitute the basic quantities (D, a and F) in the formula of time period as follows:

T = 2π√(m/k)
T = 2π√(D × V/k)
where V is the volume of the object in motion.

We know that force (F) is given by:
F = ma
where m is the mass of the object in motion and a is the acceleration.

Therefore, we can write:
a = F/m
m = D × V

Substituting these values in the formula of time period, we get:

T = 2π√(D × V/F) × √(F/m)
T = 2π√(D × V/F) × √(F/(D × V))

Simplifying the above expression, we get:

T = 2π√(D/F)


The dimensions of time period can be written as:

[T] = [D/F]^(1/2)
[T] = [M/L^3] × [L/T^2]^(-1/2)
[T] = [T]

Therefore, the dimensions of time period using density (D), acceleration (a) and force (F) as basic quantities are [T].


The dimensions of time period using density (D), acceleration (a) and force (F) as basic quantities are [T]. This means that time period is a dimensionless quantity and is not dependent on any of the given basic quantities.