__Question 2.14__:

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :

(a) y = a sin 2π t/T

(b) y = a sin vt

(c) y = (a/T) sin t/a

(d) y = (a√2) (sin 2πt / T + cos 2πt / T )

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

__Solution__:

(a)

**Answer:**Correct

y = a Sin 2πt/T

Dimension of

*y*= M

^{0}L

^{1}T

^{0}

Dimension of

*a*= M

^{0}L

^{1}T

^{0}

Dimension of Sin 2πt/T = M

^{0}L

^{0}T

^{0}

∵

Dimension of L.H.S = Dimension of R.H.S

Hence, the given formula is dimensionally correct.

(b)

**Answer:**Incorrect

*y*=

*a*sin

*vt*

Dimension of

*y*= M

^{0}L

^{1}T

^{0}

Dimension of

*a*= M

^{0}L

^{1}T

^{0}

Dimension of

*vt*= M

^{0}L

^{1}T

^{–1}× M

^{0}L

^{0}T

^{1 }= M

^{0}L

^{1}T

^{0}

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

(c)

**Answer:**Incorrect

y = (a/T) Sin(t/a)

Dimension of

*y*= M

^{0}L

^{1}T

^{0}

Dimension of a/T = M

^{0}L

^{1}T

^{–1}

Dimension of t/a = M

^{0}L

^{–1}T

^{1}

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

(d)

**Answer:**Correct

y = (a√2)(Sin 2πt/T + Cos 2πt/T)

Dimension of

*y*= M

^{0}L

^{1}T

^{0}

Dimension of

*a*= M

^{0}L

^{1}T

^{0}

Dimension of t/T = M

^{0}L

^{0}T

^{0}

^{}Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of

*y*and

*a*are the same. Hence, the given formula is dimensionally correct.

pottieee

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