Oscillations
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Question : 7
Total: 25
The motion of a particle executing simple harmonic motion is described by the displacement function.
x ( t ) = A c o s ( ω t + ϕ )
If the initial (t = 0) position of the particle is 1 cm and its initial velocity isω cm s – 1 π s – 1 x = B sin ( ω t + α )
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is
Solution:
Here, at t = 0 , x = 1 v = ω cm s – 1 ; ω = π s – 1
Given :x = A c o s ( ω t + ϕ )
Since att = 0 , x = 1
1 = A cos ( π × 0 + ϕ ) = A cos ϕ
The instantaneous particle velocity is given by
v =
=
[ A cos ( ω t + ϕ ) ] v = − A ω sin ( ω t + ϕ )
Since att = 0 , v = ω cm s – 1
ω = – A ω sin ( π × 0 + ϕ )
orA sin ϕ = – 1
Squaring and adding the equations (i) and (ii), we get
A 2 cos 2 ϕ + A 2 sin 2 ϕ = 1 2 + ( − 1 ) 2
A 2 ( cos 2 ϕ + sin 2 ϕ ) = 2 A 2 ( 1 ) = 2
orA = √ 2 c m
Dividing the equation (ii) by (i), we get
=
= − 1 ϕ =
When sine function is used to describe the simple harmonic motion.Here,x = B sin ( ω t + α )
Since att = 0 , x = 1
1 = B sin ( π × 0 + α ) B sin α = 1
The instantaneous particle velocity is given by
v =
=
[ B sin ( ω t + α ) ]
v = B ω cos ( ω t + α )
Since att = 0 , v = ω cm s – 1
ω = B ω cos ( π × 0 + α )
orB cos α = 1
Squaring and adding the equations (iii) and (iv), we get
B 2 sin 2 α + B 2 cos 2 α = 1 2 + 1 2
orB 2 ( sin 2 α + cos 2 α ) = 2
orB 2 ( 1 ) = 2
orB = √ 2 cm
Dividing the equation (iii) by (iv), we get
=
tan α = 1 α =
Given :
Since at
The instantaneous particle velocity is given by
Since at
or
Squaring and adding the equations (i) and (ii), we get
or
Dividing the equation (ii) by (i), we get
When sine function is used to describe the simple harmonic motion.Here,
Since at
The instantaneous particle velocity is given by
Since at
or
Squaring and adding the equations (iii) and (iv), we get
or
or
or
Dividing the equation (iii) by (iv), we get
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