Check dimensional analysis of T=1/l√(f/m)
Answers
Answer:
The given relation is accurate.
Step-by-step explanation:
The given relation is
v=\frac{1}{2l}\sqrt{\frac{F}{m}}v=
2l
1
m
F
Where,
v is the frequency
m is mass per unit length
F is force
l is length
If the equation is correct, the dimension on LHS should be equal to the dimensions on RHS
We know that
Dimensions of frequency = [T⁻¹]
Dimensions of length = [L]
Dimensions of Force = [MLT⁻²]
Dimensions of mass per unit length = [ML⁻¹]
Therefore,
Dimensions of LHS in the formula = [T⁻¹]
Dimensions of RHS in the formula
=\frac{1}{\text{Dimension of Length}}\sqrt{\frac{\text{Dimension of Force}}{\text{Dimension of Mass per unit Length}}}=
Dimension of Length
1
Dimension of Mass per unit Length
Dimension of Force
=\frac{1}{[L]}\sqrt{\frac{[MLT^{-2}]}{[ML^{-1}]}}=
[L]
1
[ML
−1
]
[MLT
−2
]
=\frac{1}{[L]}\sqrt{L^2T^{-2}}=
[L]
1
L
2
T
−2
=\frac{1}{[L]}\times[LT^{-1}]=
[L]
1
×[LT
−1
]
=[T^{-1}]=[T
−1
]
Thus,
Dimensions of LHS = Dimensions of RHS
Therefore, the relation is correct.
Hope this answer is helpful.