Physics, asked by drashtivadhvania15, 10 hours ago

A particle of mass 1 kg is performing SHM and its displacement any time is given by

What will be kinetic energy of particle at t = 2 sec?​

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Answers

Answered by pulakmath007
7

SOLUTION

GIVEN

A particle of mass 1 kg is performing SHM and its displacement any time is given by

 \displaystyle \sf{y = 2 \sin \bigg( \:    \frac{\pi t}{12} +  \frac{\pi}{6}  \: \bigg)}

TO DETERMINE

The kinetic energy of particle at t = 2 sec

EVALUATION

Here the given equation is

 \displaystyle \sf{y = 2 \sin \bigg( \:    \frac{\pi t}{12} +  \frac{\pi}{6}  \: \bigg)}

Comparing with the general equation

 \displaystyle \sf{y = a \sin (  \omega t +  \phi)}

We get

 \displaystyle \sf{a = 2   \:  ,\: \:     \omega = \frac{\pi }{12}  \:  ,\:  \phi =    \frac{\pi}{6}  \:}

Now for t = 2 we have

 \displaystyle \sf{y = 2 \sin \bigg( \:    \frac{2\pi }{12} +  \frac{\pi}{6}  \: \bigg)}

 \displaystyle \sf{ \implies \: y = 2 \sin \bigg( \:    \frac{\pi }{6} +  \frac{\pi}{6}  \: \bigg)}

 \displaystyle \sf{ \implies \: y = 2 \sin \bigg( \:    \frac{\pi }{3}   \: \bigg)}

 \displaystyle \sf{ \implies \: y = 2  \times  \frac{ \sqrt{3} }{2} }

 \displaystyle \sf{ \implies \: y =  \sqrt{3}  }

Hence the required kinetic energy

 \displaystyle \sf{  =  \frac{1}{2}m { \omega}^{2} ( {a}^{2} -  {y}^{2}  )  \:  \: J }

 \displaystyle \sf{  =  \frac{1}{2} \times 1 \times  {  \bigg( \frac{\pi}{12}  \bigg)}^{2} \times   \bigg[ {2}^{2} -  {( \sqrt{3}) }^{2}  \bigg]   \:  \: J }

 \displaystyle \sf{  =  \frac{1}{2} \times  \frac{ {\pi}^{2} }{144}  \times   \bigg[ 4 - 3 \bigg]   \:  \: J }

 \displaystyle \sf{  =   \frac{ {\pi}^{2} }{288} \:  \: J }

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