The amplitude of an oscillating simple pendulum is $10 \ cm$ and its period is $4 \ s$. Its speed $1 \ s$ after it passes its equilibrium position is ... $m/s$.

$A$ simple pendulum with a bob of mass $m$ and length $x$ is held in position at an angle $\theta_1$ and then at an angle $\theta_2$ with the vertical. When released from these positions,the speeds with which it passes the lowest position are $v_1$ and $v_2$ respectively. Then,the ratio $\frac{v_1}{v_2}$ is .............

$A$ simple pendulum is made of a body which is a hollow sphere containing mercury, suspended by means of a wire. If a little mercury is drained off, the period of the pendulum will

$A$ ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :

$A$ pendulum has a length of $0.4 \ m$ and a maximum speed of $4 \ m/s$. When the string makes an angle of $30^{\circ}$ with the horizontal,its speed will be: $\left[\sin \frac{\pi}{6} = \cos \frac{\pi}{3} = 0.5 \text{ and } g = 10 \ m/s^{2}\right]$

$A$ simple pendulum of length $1 \,m$ is freely suspended from the ceiling of an elevator. The time period of small oscillations as the elevator moves up with an acceleration of $2 \,m/s^2$ is (use $g=10 \,m/s^2$).