Question

If the period of S.H.M. is 12 seconds and amplitude 10 cm. What is the displacement 14 Seconds after passage of the particle through its extreme positive displacement?

• A. 5 cm
• B. 4 cm
• C. 2 cm
• D. 10 cm

Answer 1

The correct option is A 5 cm

The motion starts from extreme position, $y=A\cos \omega t=A\cos \left(\frac{2\pi }{T}\right)t$
$=10\cos [\left(\frac{2\pi }{12}\right)\times 14]=10\cos \frac{7\pi }{3}$
$=10\cos \frac{\pi }{3}=10\times \frac{1}{2}=5cm.$
Position from mean position is 5 cm hence the total displacement from the positive extreme position is 5 cm