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

College Physics
11th Edition
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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
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
Step 1

From the given problem,

fn=n2L Tμ

fn=n2L k(L-Lo)ML

fn=n2LkM1-LoL

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.

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