Two oscillating systems; a simple pendulum and a vertical spring-mass-system have same time period of motion on the surface of the Earth. If both are taken to the moon, then-
Time period of the simple pendulum will be more than that of the spring-mass system.
Time period of the simple pendulum will be equal is that is of the spring-mass system.
Time period of the simple pendulum will be less than of the spring-mass system.
Nothing can be said definitely without observation.
A mass hangs from a spring and oscillates vertically. The top end of the spring is attached to the top of a box, and the box is placed on a scale, as shown in the figure. The reading on the scale is largest when the mass is
A $2\, Kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi ^2 N/m$ attached to $2\, Kg$ block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be ... $\sec$
Block $A$ is hanging from a vertical spring and it is at rest. Block $'B'$ strikes the block $'A'$ with velocity $v$ and stick to it. Then the velocity $v$ for which the spring just attains natural length is:
What is the period of small oscillations of the block of mass $m$ if the springs are ideal and pulleys are massless ?
A mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ gives the time-period ${t_1}$ and ${t_2}$ respectively. If same mass $m$ is connected by both springs as shown in figure then time-period $t$ is given by the relation