Question

If a particle executes SHM with angular velocity $3.5 \,rad\, s^{-1}$ and maximum acceleration $7.5\, ms^{-2},$then the amplitude of oscillation will be

• A. 0.80m
• B. 0.69 m
• C. 0.41 m
• D. 0.61 m

Answer 1

Answer: (D)

0.61 m

Answer 2

The maximum acceleration for a particle executing SHM is
$\, \, \, \, \, \, \, a_{max} =\omega^2 x$
$\, \, \, \, \, \, \, \, \, \, \, x=\frac{a_{max}}{\omega^2}=\frac{7.5}{(3.5)^2}=0.61 m$