Question

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

• A. $0.53cm$
• B. $0.28m$
• C. $0.61m$
• D. $0.36m$

Answer 1

Answer: (C)

$0.61m$

$\omega =3.5$ radian/sec
Maximum acceleration of a particle under $SHM={\omega }^{2}a$ where $a$ is amplitude of oscillation.
${\omega }^{2}a=7.5\Rightarrow (3.5{)}^{2}a=7.5$

$\Rightarrow a=\frac{7.5}{3.5\times 3.5}\Rightarrow =\frac{30}{49}=0.61m$