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The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1.  At, t=0 the displacement is 5 m.  What is the maximum acceleration ?  The initial phase IS \frac{\pi }{4}
Option: 1 500 m/s2  
Option: 2 500\sqrt{2} \ m/s2
Option: 3 750 m/s2  
Option: 4750\sqrt{2} \ m/s2 m/s2  

Answers (1)


As learnt in 


f_{max}=\overset{2}{\omega}A, \:\:\:V_{max}=a \omega


X=a\: \sin \left (\omega t+ \frac {\pi}{4}\right ) [given]\:\:\:\:\:\:\:\:\:\:[omega =10]

At f=0

5=a\:\sin \frac{\pi}{4}                                        a= 5 \sqrt2

Maximum acceleration = \omega^{2} \times a

=100 \times 5 \sqrt 2

=500 \sqrt 2


Posted by

vishal kumar

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