The bob of simple pendulum of length ' L ' is released from a position of small angular displacement \theta. Its linear displacement at time ' {\t} ' is ( {\g}= acceleration due to gravity)

Correct Answer is : Lθcos[√gL⋅t]

\L \theta \cos [\sqrt{\frac{g}{L}} \cdot t]

\L \theta \sin [2 \pi \sqrt{\frac{g}{L}} \cdot t]

\L \theta \cos [2 \pi \sqrt{\frac{g}{L}} \cdot t]

\L \theta \sin [\sqrt{\frac{g}{L}} \cdot t]

MHTCET 14 May 2023 Shift 1 Solved Paper

Section: Physics

Question : 135 of 150

Marks: +1, -0

The bob of simple pendulum of length ' L ' is released from a position of small angular displacement θ. Its linear displacement at time ' t ' is ( g= acceleration due to gravity)

Lθ‌cos[√‌
g
L
⋅t]
Lθsin‌[2π√‌
g
L
⋅t]
Lθ‌cos[2π√‌
g
L
⋅t]
Lθsin‌[√‌
g
L
⋅t]

Validate

Go to Question:

Prev Question

Next Question