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Q13OQ

Expert-verifiedFound in: Page 150

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**The top end of a spring is held fixed. A block is hung on the bottom end as in Figure OQ15.13a, and the frequency f of the oscillation of the system is measured. The block, a second identical block, and the spring are carried up in a space shuttle to Earth orbit. The two blocks are attached to the ends of the spring. The spring is compressed without making adjacent coils touch (Fig. OQ15.13b), and the system is released to oscillate while floating within the shuttle cabin (Fig. OQ15.13c). What is the frequency of oscillation for this system in terms of f? (a)$\frac{f}{2}$ **

Option (d) is correct answer for this question, since we got $f\text{'}=f\sqrt{2}$

Frequency of oscillation for a particle is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

$f=$Frequency of oscillation

$k=$Spring constant

$m=$Mass of object

** **

- Let’s assume the coils of the spring do not hit one another.
- The coil in the center of the spring does not move, when the spring with two blocks is set into oscillation in space; Let’s imagine clamping the center coil in place without affecting the motion.
- Let’s duplicate the motion of each individual block in space by hanging a single block on a half-spring here on Earth.
- The half-spring with its center coil clamped or its other half cut off has twice the spring constant as the original uncut spring because an applied force of the same size would produce only one half the extension distance.
- Thus the oscillation frequency in space is$f\text{'}=\frac{1}{2\pi}\sqrt{\frac{k\text{'}}{m}}=\frac{1}{2\pi}\sqrt{\frac{2k}{m}}=f\sqrt{2}$ . The absence of a force required to support the vibrating system in orbital free fall has no effect on the frequency of its vibration.
- Hence option (d) is correct answer for this question.

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