Question

A 2.5 kg collar attached to a spring of force constant 1000 N m-1 slides without friction on a horizontal rod as shown in the figure. The collar is displaced from its equilibrium position by 5.0 cm and released. Calculate (1) the period of oscillation (2) the maximum magnitude of acceleration and (3) the maximum speed of the collar.​

Answer 1

Answer:

Maximum speed,$2 m s^{-1}$,The speed of oscillation, 0.63 s.

Step-by-step explanation:

Stage 1: We might compute the period and, in this way, the recurrence by estimating the length of one full wavering. Know that the time of a pendulum is free of its mass, however the time of a mass on a spring is autonomous of the spring's length.

Stage 2: The most elevated pace an individual can reach is called greatest speed. At the point when competitors modify the greatness (how rapidly they are moving) or bearing (or both) of their movement, or both, their speed increase changes too since speed increase is connected with speed, which has both an extent and a heading associated with it.

Step 3: The speed of oscillation , $T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{2.5}{250}}=\frac{2 \times 3.14}{10}=0.63 \mathrm{~s}$

Maximum speed,

$v_m=A \omega=0.2 \sqrt{\frac{k}{m}}=0.2 \times \sqrt{\frac{250}{2.5}}=0.2 \times 10=2 \mathrm{~ms}^{-1}$

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