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

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asked Feb 22, 2018 in Physics by anukriti (13,536 points) 5 10 19
edited Feb 22, 2018 by anukriti

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

y = asin(2πt/T)

y = a sin vt

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

y = (a2)(sin2πt/T + cos2πt/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.

1 Answer

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answered Feb 22, 2018 by Ankit Agarwal (28,847 points) 7 31 67
selected Feb 22, 2018 by anukriti
 
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Correct y = asin(2πt/T)

Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimension of sin(2πt/T) = M0 L0 T0
Dimension of L.H.S = Dimension of R.H.S
Hence, the given formula is dimensionally correct.
Incorrect y = a sin vt
Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimension of vt = M0 L1 T–1 × M0 L0 T1 = M0 L1 T0

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.

Incorrect y = (a/T) sin(t/a)
Dimension of y = M0L1T0
Dimension of a/T = M0L1T–1
Dimension of t/a = M0 L–1 T1

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

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.

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