Chapter 43 of 2363 min read

Spring–mass system and simple pendulum

Why this is asked: A spring-mass system's period, T = 2π√(m/k), doesn't involve g and works the same on the Moon, while a pendulum's period, T = 2π√(L/g), depends on effective g and simply stops swinging in a freely falling lift.

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Test yourself

•In an oscillating spring–mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $ω (t)$ and average amplitude $A (t)$ of the system change with time $t$ . Which one of the following options (graphs shown) schematically depicts these changes correctly?
•Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_{1}$ and $k_{2}$ respectively, oscillate vertically. If their maximum speeds are the same, the ratio $A_{Q} / A_{P}$ of the amplitude $A_{Q}$ of mass Q to the amplitude $A_{P}$ of mass P is:
•If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:

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