Question

A student was asked to write the equation for displacement at any instant in a simple harmonic motion of amplitude ‘a’. He wrote the equation as y = a sin (2πvt/k)
Where ‘v’ is the velocity at instant t sec. For the equation to be dimensionally correct, what should be the dimensions of k?​

Answer 1

Answer: Y=asin 2πvt/K

[Y] = [L] [LT - 1 ] [T] / [K]

[L] = [L 2] / [K]

[K] = [L]

Answer 2

For the equation to be dimensionally correct, the dimensions of 'k' should be equal to [L].

Step-by-step Explanation:

In the equation of the simple harmonic wave:     $Y = asin(\frac{2\pi vt}{k} )$

Dimension of displacement (Y) = [L]

Dimension of amplitude (a) = [L]

Dimension of velocity (v) = [LT⁻¹]

Dimension of time (t) = [T]

→ The arguments of trigonometric functions such as sine are always a dimensionless quantity because the argument has to be a purely real number. Therefore the 'θ' inside the sinθ must be dimensionless.

→ Hence, 2πvt/k must be dimensionless.

→Let the dimensions of 'k' be [Mᵃ Lᵇ Tⁿ].

                                   $\frac{[LT^{-1}][T]}{[M^aL^bT^n]} =[M^0L^0T^0]\\\\\[[M^{-a} L^{1-b} T^{-n} ]=[M^0L^0T^0]$

        ∵ -a = 0                             ∵ 1 - b = 0                          ∵ -n = 0

         ∴ a = 0                                ∴ b = 1                              ∴ n = 0

→ The dimension of 'k' comes out to be equal to [M⁰ L T⁰].

Therefore, for the equation to be dimensionally correct, the dimensions of 'k' should be equal to [L], which is the dimension of length.

#SPJ2