Question
Question: A particle executes simple harmonic motion and is located at \( x = a,b \) and \( c \) at times \( {...
A particle executes simple harmonic motion and is located at x=a,b and c at times t0,2t0 and 3t0 respectively. The frequency of the oscillation is
(A) 2πt01cos−1(b2a+3c)
(B) 2πt01cos−1(2ba+c)
(C) 2πt01cos−1(2ca+c)
(D) 2πt01cos−1(2ca+2b)
Solution
Hint : to solve this problem we should know about simple harmonic motion. The general SHM equation applies to all simple oscillating motion is,
x=x0cos(ωt)
Here, x0 is the amplitude of the SHM and ω is the angular frequency of the SHM.
Complete Step By Step Answer:
A particle executes simple harmonic motion having A is amplitude of SHM and ω is angular frequency of the SHM.
As per location given in question. Equation of simple harmonic motion will be at x=a,b and c at given time t0,2t0 and 3t0 will be respectively,
a=Acosωt0 ………………………… (1)
b=Acos2ωt0 ………………………… (2)
c=Acos3ωt0 ………………………… (3)
On adding (1) and (3) . We get,
a+c=A(cosωt0+cos3ωt0)
By applying a trigonometric equation.
⇒a+c=2A(cos(23ωt0+ωt0)cos(23ωt0−ωt0))
⇒a+c=2Acos2ωt0cosωt0
From (2) we get,
b=Acos2ωt0
⇒a+c=2bcosωt0
By taking the inverse. We get,
⇒cos−1(2ba+c)=ωt0
As we know ω=2πf . So,
⇒cos−1(2ba+c)=2πft0
⇒f=2πt01cos−1(2ba+c)
Hence, (b) 2πt01cos−1(2ba+c) is correct option.
Note :
In simple harmonic motion particles oscillate about their mean position about which particle is to its to and for motion. In simple harmonic motion distance is directly proportional to acceleration of the particle. The maximum kinetic energy of a particle is at mean position and maximum potential energy is at maximum position and vice-versa.