Question

An object moves vertically with simple harmonic motion just behind a wall. From the other side of the wall the object is visible in each cycle for 2.0s and hidden behind the wall for 6.0s The maximum height reached by the object relative to the top of the wall is 0.3 m. The amplitude of the motion is :

• A. 0.5 m
• B. 0.6 m
• C. 1.0m
• D. 1.2m

Answer 1

Answer: (C)

1.0m

Answer 2

Time period of motion = 6 + 2 = 8s

from mean position to the highest point of the wall, it takes 1s

and covers distance $\frac{A}{\sqrt{2}}$

Thus A - $\frac{A}{\sqrt{2}}$ = 0.3 m

⇒ A = 1.0 m