Question

T₁ is the time period of simple pendulum. The point of suspension moves vertically upwards according to y = kt², where k = 1 m/s². New time period is T₂, then \(\frac{{T_1^2}}{{T_2^2}} = ?\)  (g = 10 m/s²)
1. 4/5
2. 6/5
3. 5/6
4. 1

Answer 1

Correct Answer - Option 2 : 6/5

Acceleration of the point of suspension

\(\begin{array}{l} a\; = \;\frac{{{d^2}y}}{{d{t^2}}}\; = \;2k\; = \;2\;m/{s^2}\\ T\; = \;2\pi \sqrt {\frac{L}{{{g_{eff}}}}} \end{array}\)

For simple pendulum: \({T_1}\; = \;2\pi \sqrt {\frac{L}{{10}}}\)

When point of suspension moves vertically upwards; g_eff = 10 + 2 = 12 m/s²

\(\begin{array}{l} {T_2}\; = \;2\pi \sqrt {\frac{L}{{12}}} \\ \therefore \frac{{T_1^2}}{{T_2^2}}\; = \;\frac{6}{5} \end{array}\)