Question

The velocity of a particle in SHM at the instant when it is $0.6$ cm away from the mean position is $4$ cm /sec. If the amplitude of vibration is $1$ cm then its velocity at the instant when it is $0.8$cm away from the mean position is:

• A. $3.0$ cm/s
• B. $2.5$ cm/s
• C. $2.25$ cm/s
• D. $3.5$ cm/s

Answer 1

Answer: (A)

$V=\omega \sqrt{A^2-x^2}$

$V_1=\omega \sqrt{A^2-x^2}$

$V_1=\omega \sqrt{A^2-(0.6{)}^2}$

$V_2=\omega \sqrt{A^2-x^2}$

$V_2=\omega \sqrt{A^2-(0.8{)}^2}$

$\frac{V_1}{V_2}=\frac{\sqrt{(1^2-(0.6{)}^2}}{\sqrt{(1^2-(0.8{)}^2}}$

$\frac{4}{V_2}=\frac{0.8}{0.6}$

$V_2=3cm/sec$