QJEEM07Ph9dba: JEE Main 2007 Physics - Class 11 JEEM07Ph9dba Solution: A point mass oscillates along the -axis according to the law If …

Question

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A point mass oscillates along the

x

-axis according to the law

x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).

If the acceleration of the particle is written as

a = A\,\cos \left( {\omega t + \delta } \right),

then

Accepted Answer

Here,

x = {x_0}\cos \left( {\omega t - \pi /4} \right)

\therefore

Velocity,

v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)

Acceleration,

a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)

= {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]

= {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)

\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Acceleration,

a = A\cos \left( {\omega t + \delta } \right)

\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Comparing the two equations, we get

A = {x_0}{\omega ^2}

and

\delta = {{3\pi } \over 4}.

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