Given that the displacement of an oscillating particle is given by y = A sin (Bx + Ct + D). The dimensional formula for (ABCD) is

Question

Given that the displacement of an oscillating particle is given by y = A sin (Bx + Ct + D). The dimensional formula for (ABCD) is (A)  [ M0L-1T0 ]  (B)  [ M0L0T-1 ]  (C)  [ M0L-1T-1 ]  (D)  [ M0L0T0 ]

Tardigrade Admin · Accepted Answer

y=A sin (Bx+Ct+D) As each term inside the bracket is dimensionless, so A=y=[L1] B=(1/x)=[L-1] C=(1/t)=[T-1] and D is dimensionless. ∴ [ABCD]=[L][L-1][T-1][1] =[M0L0T-1]

Q. Given that the displacement of an oscillating particle is given by y = A sin (Bx + Ct + D). The dimensional formula for (ABCD) is

$[M^{0} L^{- 1} T^{0}]$

$[M^{0} L^{0} T^{- 1}]$

$[M^{0} L^{- 1} T^{- 1}]$

$[M^{0} L^{0} T^{0}]$

$y = A sin (B x + Ct + D)$ As each term inside the bracket is dimensionless, so $A = y = [L^{1}]$ $B = \frac{1}{x} = [L^{- 1}]$ $C = \frac{1}{t} = [T^{- 1}]$ and D is dimensionless. $∴$ $[A BC D] = [L] [L^{- 1}] [T^{- 1}] [1]$ $= [M^{0} L^{0} T^{- 1}]$

Q. Given that the displacement of an oscillating particle is given by y = A sin (Bx + Ct + D). The dimensional formula for (ABCD) is

[M0L−1T0]

[M0L0T−1]

[M0L−1T−1]

[M0L0T0]

y=Asin(Bx+Ct+D) As each term inside the bracket is dimensionless, so A=y=[L1] B=x1​=[L−1] C=t1​=[T−1] and D is dimensionless. ∴ [ABCD]=[L][L−1][T−1][1] =[M0L0T−1]

$[M^{0} L^{- 1} T^{0}]$

$[M^{0} L^{0} T^{- 1}]$

$[M^{0} L^{- 1} T^{- 1}]$

$[M^{0} L^{0} T^{0}]$

$y = A sin (B x + Ct + D)$ As each term inside the bracket is dimensionless, so $A = y = [L^{1}]$ $B = \frac{1}{x} = [L^{- 1}]$ $C = \frac{1}{t} = [T^{- 1}]$ and D is dimensionless. $∴$ $[A BC D] = [L] [L^{- 1}] [T^{- 1}] [1]$ $= [M^{0} L^{0} T^{- 1}]$