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  • A particle is performing simple harmonic motion along the axis with amplitude 4 cm and time period

A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x =2 cm to x = + 4 cm  and back again is given by

A)            0.6 sec                                     

B)            0.4 sec

C)            0.3 sec                                     

D)            0.2 sec

 

Answers (2)

Time taken by particle to move from x=0 (mean position) to x = 4 (extreme position)

=T/4=1.2/4=0.3 s 

Let t be the time taken by the particle to move from x=0 to x=2 cm                    

\\y=asin\omega t\Rightarrow 2=4sin2\pi t/T\\\\1/2=sin2\pi t /1.2\\\\\frac{\pi}{6}=\frac{2\pi }{1.2}t\\\\t=0.1s

Hence time to move from x = 2 to x = 4 will be equal to 0.3 ?

0.1 = 0.2 s           

Hence total time to move from x = 2 to x = 4 and back again 

=2\times0.2=0.4sec

Posted by

lovekush

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Time taken by particle to move from x=0 (mean position) to x = 4 (extreme position)

=T/4=1.2/4=0.3 s 

Let t be the time taken by the particle to move from x=0 to x=2 cm                    

\\y=asin\omega t\Rightarrow 2=4sin2\pi t/T\\\\1/2=sin2\pi t /1.2\\\\\frac{\pi}{6}=\frac{2\pi }{1.2}t\\\\t=0.1s

Hence time to move from x = 2 to x = 4 will be equal to 0.3 ?

0.1 = 0.2 s           

Hence total time to move from x = 2 to x = 4 and back again 

=2\times0.2=0.4sec

Posted by

lovekush

View full answer

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