Question

Displacement time equation of a particle executing SHM is $x=10sin(\frac{\pi }{3}t+\frac{\pi }{6})cm$. The distance covered by the particle in 3 seconds is

• A. 5 cm
• B. 20 cm
• C. 10 cm
• D. 15 cm

Answer 1

Answer: (B)

20 cm

Given,

$A=10cm$

$\omega =\frac{2\pi }{T}$

$T=\frac{2\pi }{\omega }=6s$

i.e mean to extreme it takes $1.5s.$

Hence in $3s$, it will travel twice the amplitude, making a net zero displacement,

but covering a $2A$ distance.

i.e $20cm$

Option $\text{B}$ is the correct answer