A particle executing SHM while moving from one extremity is found at distance x1, x2 and x3 from the centre at the end of three successive seconds. The time period of oscillation is where θ= cos -1((x1+x3/2 x2))

Question

A particle executing SHM while moving from one extremity is found at distance x1, x2 and x3 from the centre at the end of three successive seconds. The time period of oscillation is where θ= cos -1((x1+x3/2 x2)) (A) 2 π /θ (B) π/ θ (C) θ (D) π / 2 θ

Tardigrade Admin · Accepted Answer

Displacement-time equation of the particle will be

X = A cos ω t

Given that

x_{1} = A cos ω

and

x_{2} = A cos 2 ω

Now,

\frac{x _{1} + x _{3}}{2 x _{2}} = \frac{A ( c o s θ + c o s 3 ω )}{2 A c o s 2 ω}

= \frac{2 A c o s 2 ω c o s ω}{2 A c o s 2 ω}

= cos ω

ω = cos^{- 1} (\frac{x _{1} + x _{2}}{2 x _{2}}) = \frac{2 π}{T}

or

T = \frac{2 π}{θ}

where

θ = cos^{- 1} (\frac{x _{1} + x _{3}}{2 x _{2}})

Q. A particle executing SHM while moving from one extremity is found at distance $x_{1}, x_{2}$ and $x_{3}$ from the centre at the end of three successive seconds. The time period of oscillation is where $θ = cos^{- 1} (\frac{x _{1} + x _{3}}{2 x _{2}})$

$2 π / θ$

$π / θ$

$θ$

$π /2 θ$

Q. A particle executing SHM while moving from one extremity is found at distance x1​,x2​ and x3​ from the centre at the end of three successive seconds. The time period of oscillation is where θ=cos−1(2x2​x1​+x3​​)

2π/θ

π/θ

θ

π/2θ

Displacement-time equation of the particle will be X=Acosωt Given that x1​=Acosω and x2​=Acos2ω Now, 2x2​x1​+x3​​=2Acos2ωA(cosθ+cos3ω)​ =2Acos2ω2Acos2ωcosω​ =cosω ω=cos−1(2x2​x1​+x2​​)=T2π​ or T=θ2π​ where θ=cos−1(2x2​x1​+x3​​)

Q. A particle executing SHM while moving from one extremity is found at distance $x_{1}, x_{2}$ and $x_{3}$ from the centre at the end of three successive seconds. The time period of oscillation is where $θ = cos^{- 1} (\frac{x _{1} + x _{3}}{2 x _{2}})$

$2 π / θ$

$π / θ$

$θ$

$π /2 θ$