check the correctness of equation y = a sin 2pit/T where y is displacement,a is maximum displacement ,t is time at any instant and T is time period
Answers
Answer:
Since all the equations are for the displacements of the particle, they all should have the dimension of the length.
Now, checking all the equations of displacement to be dimensionally correct.
(a)
Since the tignometric functions are dimensionless,
T
2πt
will also be dimensionless.
T
2πt
=
T
T
=1=[M
0
L
0
T
0
]
Here, a is the amplitude and it will have the dimension of the length. SO, the equation is correct.
(b)
Since the tignometric functions are dimensionless, vt should be dimensionless.
vt=(LT
−1
)(T)=L=[M
0
L
1
T
0
]
Thus, the given equation is not correct.
(c)
The quantity
a
t
should be dimensionless
a
t
=
L
T
=[M
0
L
−1
T
1
]
Therefore, it is also dimensionally incorrect.
(d)
The tignometric functions are dimensionless,
T
2πt
will also be dimensionless.
T
2πt
=
T
T
=1=[M
0
L
0
T
0
]
The given equation is dimensionally correct.
Thus, the formulas in (b) & (c) are dimensionally wrong.
given equation--y=aSin 2π t/T
a=amplitude
y=length
t=time
T=time period
Writing dimensional formula on both side of the equation.[Note>>>constant values and trigonometric functions does not have dimensional formula.]
y=a Sin2πt/T
[L]=[L]×[T]×1/[T]
[L]=[L]×1. (T & T gets cancel)
[L]=[L]
Hence the given equation is dimensionally correct.