The time period of a simple pendulum is $2\, sec$. If its length is increased $4$ times, then its period becomes ..... $\sec$
$16$
$12$
$8$
$4$
What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10\,cm$ ($g = 9.8\, m/s^2$) ..... $m/s$
Answer the following questions:
$(a)$ Time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle:
$T=2 \pi \sqrt{\frac{m}{k}}$. A stmple pendulum executes $SHM$ approximately. Why then is the time pertodof.anondwers period of a pendulum independent of the mass of the pendulum?
$(b)$ The motion of a simple pendulum is approximately stmple harmonte for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}} .$ Think of a qualitative argument to appreciate this result.
$(c)$ A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
$(d)$ What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely failing under gravity?
The length of a seconds pendulum at a height $h=2 R$ from earth surface will be.(Given: $R =$ Radius of earth and acceleration due to gravity at the surface of earth $g =\pi^{2}\,m / s ^{-2}$ )
There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4,$ then the new time period of the pendulum is
The amplitude of an oscillating simple pendulum is $10\,cm$ and its period is $4\, sec$. Its speed after $1\, sec$ after it passes its equilibrium position, is ... $m/s$