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A Book with Many Printing Errors Contains Four Different Formulas for the Displacement Y of a Particle Undergoing a Certain Periodic Motion: - Physics

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प्रश्न

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 `(2pit)/T`

(b) y = a sin vt

(c) y = `(a/T) sin  t/a`

d) y = `(a/sqrt2) (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.

उत्तर १

a) Correct

y = `asin  (2pit)/T`

Dimension of y = M0 L1 T0

Dimension of a = M0 L1 T0

Dimension of `sin (2pit)/T = M^0L^0T^0`

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

Hence, the given formula is dimensionally correct.

b) Incorrect

Dimension of y = M0 L1 T0

Dimension of a = M0 L1 T0

Dimension of vt = M0 L1 T–1 × M0 L0 T= 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.

3) Incorrect

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

Dimension of y = M0L1T0

Dimension of `a/T` = `M^0L^1T^(-1)`

Dimension of `t/a` = `M^0L^(-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.

4)Correct

y = `(asqrt2)(sin2pi t/T + cos2pi t/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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उत्तर २

According to dimensional analysis an equation must be dimensionally homogeneous.

a) `y = a sin  (2pit)/T`

Here, [L.H.S] = [y] = [L] and [R.H.S] = `[a sin  (2pit)/T]` = `[L sin  T/T]` = [L]

So, it is correct.

b) y = a sin vt

Here [y] = [L] and `[a sin vt] = [L sin(LT^(-1) T)] = [L sinL`]

So the equation is wrong

c) `y = (a/T) sin  t/a`

Here [y] = [L] and  `[(a/T) sin  t/a] = [L/T sin  T/L] = [LT^(-1) sin TL^(-1)]`

So the equation is wrong.

d) y = `(asqrt2)(sin  (2pit)/T + cos  (2pit)/T)`

Here, [y] = [L],[`asqrt2`] = [L]

and` [sin  (2pit)/T + cos  (2pit)/T] = [sin  T/T + cos T/T]` = dimensionless

So the equation is correct

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Units and Measurements - Exercises [पृष्ठ ३६]

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एनसीईआरटी Physics [English] Class 11
अध्याय 2 Units and Measurements
Exercises | Q 14 | पृष्ठ ३६

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