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Chapter 15: Q. 50 (page 417)

A mass hanging from a spring oscillates with a period of0.35s. Suppose the mass and spring are swung in a horizontal circle, with the free end of the spring at the pivot. What rotation frequency, in rpm, will cause the spring's length to stretch by 15%?

Short Answer

Expert verified

Rotational frequency in rpm is66rpm.

Step by step solution

01

Expression for km

The period of oscillation of the mass is,

T=2蟺mk

Rearrange it,

T2=4蟺2mk

km=4蟺2T2

02

Expression for rotation frequency

A radial force acting toward the path's center of curvature is required for the mass to travel in a uniform circular motion.

In this situation, the radial force is the force exerted by the spring on the mass.

The net force on the mass is,

鈭�F=办螖谤=尘蝇2r

蝇=km螖谤r

Substitute all values,

蝇=4蟺2T2螖谤r

03

Calculation for rotational frequency

Frequency is,

蝇=4蟺2T2螖谤r

WhereT=0.35s,鈭�rr=0.15

蝇=4蟺2(0.35s)2(0.15)

=6.95rad/s

converted to rpm,

=(6.95rad/s)1revolution2蟺谤补诲60s1min

=66rpm

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