Physics, asked by venkatreddygopireddy, 2 months ago

the displacement of two identical particles executing SHM are represented equations
X1=4 sin (10t+π/6) and x2 =5cos (omega t). for what value of Omega energy of a both the particles is same​

Answers

Answered by shadowsabers03
5

If the displacement of a particle executing SHM is represented as,

\sf{\longrightarrow x=A\cdot f(\omega t+\phi)}

where \sf{f(x)=\sin x} or \sf{f(x)=\cos x,} then the total energy of the particle will be,

\sf{\longrightarrow E=\dfrac{1}{2}\,m\omega^2A^2}

Here the particles are identical, so,

\sf{\longrightarrow m_1=m_2=m}

The displacement of the first particle is represented as,

\sf{\longrightarrow x_1=4\sin\left(10t+\dfrac{\pi}{6}\right)}

Here,

  • \sf{A_1=4}
  • \sf{\omega_1=10}

Then, total energy,

\sf{\longrightarrow E_1=\dfrac{1}{2}\,m_1\left(\omega_1\right)^2\left(A_1\right)^2}

\sf{\longrightarrow E_1=800m}

The displacement of the second particle is represented as,

\sf{\longrightarrow x_2=5\cos\left(\omega t\right)}

Here,

  • \sf{A_2=5}
  • \sf{\omega_2=\omega}

Then, total energy,

\sf{\longrightarrow E_2=\dfrac{1}{2}\,m_2\left(\omega_2\right)^2\left(A_2\right)^2}

\sf{\longrightarrow E_2=\dfrac{25}{2}\,m\omega^2}

Now,

\sf{\longrightarrow E_1=E_2}

\sf{\longrightarrow800m=\dfrac{25}{2}\,m\omega^2}

\sf{\longrightarrow\omega^2=64}

\sf{\longrightarrow\underline{\underline{\omega=\pm8}}}

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