| DIRECTION: Each of the questions contains two statements: Statements-1 (Assertion) and Statements-2 (Reason). Choose the correct answer (Only one option is correct) from the following- |
| Statement-1: In simple harmonic motion, the velocity is maximum when the acceleration is minimum. |
| Statement-2: Displacement and velocity of SHM differ in phase by \[\frac{\pi }{2}.\] |
A) Statements-1 is false, Statements-2 is true.
B) Statements-1 is true, Statements-2 is true; Statements-2 is a correct explanation for Statements-1.
C) Statements-1 is true, Statements-2 is true; Statements-2 is not a correct explanation for Statements-1.
D) Statements-1 is true, Statements-2 is false.
Correct Answer: B
Solution :
At the middle point velocity of the particle under SHM is maximum but acceleration is zero since displacement is zero. So statement is true. We know that x= a sin \[\omega t\] ...(1) where x is displacement & a is amplitude. Velocity \[=\frac{dx}{dt}=a\omega \cos \omega t\] \[=a\omega \cos (-\omega t)=a\omega \sin \left( \frac{\pi }{2}-(-\omega t) \right)\] \[=a\omega \sin \left( \omega t+\frac{\pi }{2} \right)\] ?(2) From (i) &, (ii), it is clear that velocity is ahead of displacement (x) by \[\frac{\pi }{2}\]angle. Now, \[v=a\omega \cos \omega t\] \[a=\frac{dv}{dt}=-a{{\omega }^{2}}\sin \omega t\] Now, when velocity is maximum, from equation (ii) it is clear that \[\omega t+\frac{\pi }{2}=\frac{\pi }{2}\Rightarrow \omega t=0\] If we put the value of \[\omega t=0\]in acceleration equation, it becomes \[a=\frac{dv}{dt}=-a{{\omega }^{2}}\sin 0=0\] So, when velocity is maximum, acceleration is minimum (zero). As statement-1 is based upon statement-2, so option [b] is correct.
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