
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Transcribed Image Text:For a spring that obeys Hooke's law, the tension in the spring is proportional to the stretched length
of the spring T = k(L – Lo), where L and L, are the length of the stretched and unstretched
spring, respectively, and k is the spring constant. The mass density of the spring is μ = M/L and
the wave velocity is v = √√√//μ. Substitution of these expressions into Eq. 6.1 shows that the
frequency of the standing waves depends on the length of the spring Las
=√√√√1-2.
You will use this equation in Sec. 6.5 of the experiment.
For a slinky, the unstretched length is much shorter than the stretched length. Suppose that the
fundamental standing wave is oscillating on a slinky. How does the period of the standing wave
change as the lengthof the slinky is altered?
A. It hardly changes
B. It gets longer
C. It gets shorter
Expert Solution

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Step 1
From the given problem,
Here, the upstretched length is much shorter than the stretched length.
i.e Lo << L
From the above comparison, we can see that
The change in length of slinky will not affect the frequency as Lo << L.
Therefore, there is no change in time period.
Step by stepSolved in 2 steps

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