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.
NCERT Solutions for Class 11 Physics Part 1 Chapter 2 Exercise 14
Answers
t : time = T
Time period = T
velocity : L / T
a = displacement = L
y = displacement = L
a) L = L Sin L/L this is okay...
b) L = L sin L/T * T = L Sin L ... this is not alright, as L is not radians.. Argument of Sine function should be in radians and dimensionless.
c) L = L/T Sin T/L not correct
d) L = L (Sin T/T + Cos T/T) this is alright...
Answer:
y = Dimension of the length = [L]
v = [L]/[T]
t = [T]
a = [L]
Now let's check whether the given formulas are correct.
Note: If R.H.S = [L] then the given formula is dimensionally correct.
(a) y = a sin 2π t/T ( Since the tignometric functions are dimensionless, sin2π will also be dimensionless.)
y = [L] [T] /[T]
y = [M0L1T0]
[L] = [L]
L.H.S = R.H.S
The equation is dimensionally correct.
(b) y = a sin vt ( Since the tignometric functions are dimensionless, sin will also be dimensionless.)
y = [L] [L][T] [T^-1]
y = [M0L^2T0]
y = [L^2]
but y = [L]
The given equation is not dimensionally correct.
(c) y = (a/T) sin t/a ( Since the tignometric functions are dimensionless, sin will also be dimensionless.)
y = [L]/[T] * [T]/[L]
y = [M0L0T0]
but y = [L]
The given equation is not dimensionally correct.
(d) y= (a√2) (sin 2πt / T + cos 2πt / T )
( Since the tignometric functions are dimensionless, sin2π,cosπ will also be dimensionless.)
y = [L] ([T]/[T])+ ([T]/[T])
y = [L] [T^0]
y = [L^1M0T0]
y = [L]
[L] = [L]
L.H.S = R.H.SThe equation is dimensionally correct.
Thus , only (a) and (d) are dimensionally correct.
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