 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q13OQ

Expert-verified Found in: Page 150 ### Physics For Scientists & Engineers

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

See the step by step solution

## Step 1: Frequency of oscillation

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

## Step 2: Find the frequency of oscillation for this system

• 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. ### Want to see more solutions like these? 