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Chapter 1: Q80P (page 1)

A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

Short Answer

Expert verified

(a) 4.05 kg

Step by step solution

01

Given Data

\(\begin{aligned}{c}{\rm{force}}\;(F) = 40\;{\rm{N}}\\{\rm{distance,}}\;x = 0.25\;{\rm{m}}\\{\rm{period}},\;T = 1\;{\rm{s}}\\{\rm{Amplitude}},\;A = 0.05\;{\rm{m}}\end{aligned}\)

02

Concept

The equilibrium position means when the state of motion of a system or its internal energy never changes with time.

03

Step 3(a): Find the mass

Force in spring by Applying Hook鈥檚 law,

\(\begin{aligned}{c}\left| F \right| = kx\\40 = k\left( {0.25} \right)\\k = 160\;{\rm{N/m}}\end{aligned}\)

Time period,

\(\begin{aligned}{c}T = 2\pi \sqrt {\frac{m}{k}} \\1 = 2\pi \sqrt {\frac{m}{{160}}} \\m = \frac{{160}}{{{{\left( {2\pi } \right)}^2}}}\\ = 4.05\;{\rm{kg}}\end{aligned}\)

Hence, the mass is 4.05kg.

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