All Problems

Mechanical Oscillations

Problem 4.74

A deadweight suspended from a weightless spring extends it by Δx=9.8 cm.\Delta x=9.8 \mathrm{~cm} . What will be the oscillation period of the deadweight when it is pushed slightly in the vertical direction? The logarithmic damping decrement is equal to λ=3.1\lambda=3.1.

Reveal Answer
T=(4π2+λ2)Δx/g=0.70 sT=\sqrt{\left(4 \pi^{2}+\lambda^{2}\right) \Delta x / g}=0.70 \mathrm{~s}
Prev
Next