NCERT Solutions Class 11 Physics - Solved Exercise Question 2.14

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 = 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.

(b) Answer: 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.

(c) Answer: 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.

(d) Answer: Correct
y = (a√2)(Sin 2πt/T + Cos 2πt/T)
Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimension of t/T = M0 L0 T0

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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