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Chapter 9: Q.58 (page 230)

A spring of equilibrium length L1and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. Then a second spring, with equilibrium length L2 and spring constant k2, is hung from the bottom of m1. Mass m2 is suspended from this second spring. How far is m2 below the ceiling?

Short Answer

Expert verified

The distance between ceiling and lower attached mass is(L1+L2)[(m1+m2)k1+m2k2]g

Step by step solution

01

Content Introduction

The spring force is calculated as

Fsp=k鈭�y

Here, Fspis the spring force, kis the spring constant and 鈭�yis the change in the length of spring.

Looking at the following diagram,

The stretch in the upper spring is calculated as

鈭�y1=(m1+m2)gk1

The stretch in the lower spring is calculated as

鈭�y2=m2gk2

02

Content Explanation

The total length is calculated as follows:

Totallength=(L1+鈭�y1)+(L2+鈭�y2)=[L1+(m1+m2)gk1]+[L2+m2gk2]=(L1+L2)[(m1+m2)k1+m2k2]g

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