Question

# A particle execute simple harmonic motion with an angular velocity of $$3.5$$ rad/sec and maximum acceleration $$7.5 m/ s^2$$ . The amplitude of oscillation will be

A
0.53cm
B
0.28m
C
0.61m
D
0.36m

Solution

## The correct option is B $$0.61 m$$$$\omega = 3.5$$ radian/secMaximum acceleration of a particle under $$SHM = \omega ^ {2}a$$ where $$a$$ is amplitude of oscillation.$$\omega ^{2}a=7.5 \Rightarrow (3.5)^{2}a=7.5$$$$\displaystyle \Rightarrow a=\frac {7.5}{3.5 \times 3.5} \Rightarrow = \frac {30}{49}=0.61m$$Physics

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