Question
Question: A particle moving with simple harmonic motion has speed 3cm/s and 4cm/s at displacement 8cm and 6c...
A particle moving with simple harmonic motion has speed 3cm/s and 4cm/s at
displacement 8cm and 6cm, respectively, from the equilibrium position. Find
(a)the period of oscillation, and
(b)the amplitude of oscillation.
Solution
For a body performing SHM velocity at a given displacement is given by the equationv=ωA2−x2. We can divide the two equations for the two given displacements to get the desired variable (here amplitude A). The time period for a body performing SHM is given byT=ω2π.
Step by step answer: The velocity of a body performing simple harmonic motion at a given displacement is given by the equation v=ωA2−x2.
Given, at a displacement x= 8 cm the velocity of the body performing SHM v=3cm/s.
∴3=ωA2−82 ----------(1)
And for the displacement x=6cm, the velocity of the body v=4cm/s.
∴4=ωA2−62 ----------(2)
Dividing both equations we get,
43=ωA2−36ωA2−64 ⇒(43)2=A2−36A2−64 ⇒9A2−324=16A2−1024 ⇒7A2=700 ⇒A2=7700=100 ∴A=100=10
Therefore the amplitude of the body is 10cm.
Substituting the value of amplitude 'A' in equation 1 we get,
3=ω102−82 ⇒3=ω100−64 ⇒3=ω36 ∴ω=63=21
The time period of body performing simple harmonic motion is given by T=ω2π
Substituting the value of ω in the equation we get T=212π=4π
Therefore the time period =4π and the amplitude of the body is 10cm.
Note: For a particle performing simple harmonic motion its acceleration is directly proportional to its displacement from its mean position and is directed always opposite to its displacement. We can derive the equation of velocity from the general equation of simple harmonic motion by differentiating it.