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 sin2 - t-Tb y-a sin v tc y-a-T sint-ad y-a -2sin2 - t-T-cos2 - t-Ta- maximum displacement of the particle- v- speed of the particle- T- time-period of motion- Rule out the wrong formulas on dimensional grounds-

(a) Answer: Correct Dimension of y = M0 L1 T0 Dimension of a = M0 L1 T0 Dimension of = M0 L0 T0 Dimension of L.H.S = Dimension of R.H.S Hence, the given formu ...

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Byju's Answer

Standard XII

Physics

Distance vs Displacement

A book with m...

Question

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

(a)

(b) y = a sin vt

(c)

(d)

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

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Solution

(a) Answer: Correct

Dimension of y = M⁰ L¹ T⁰

Dimension of a = M⁰ L¹ T⁰

Dimension of = 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

Dimension of y = M⁰L¹T⁰

Dimension of = M⁰L¹T^–1

Dimension of= 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

Dimension of y = M⁰ L¹ T⁰

Dimension of a = M⁰ L¹ T⁰

Dimension of = 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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