Simple Harmonic Motion Ques 3
- A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12}$ per second. What is the force constant of the bonds connecting one atom with the other? (Take, molecular weight of silver $=108$ and Avogadro number $=6.02 \times 10^{23} \mathrm{~g} \mathrm{~mol}^{-1}$ )
(2018 Main)
(a) $5.5 \mathrm{~N} / \mathrm{m}$
(b) $6.4 \mathrm{~N} / \mathrm{m}$
(c) $7.1 \mathrm{~N} / \mathrm{m}$
(d) $2.2 \mathrm{~N} / \mathrm{m}$
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Answer:
Correct Answer: 3.( c )
Solution:
- Given, frequency, $f=10^{12} / \mathrm{sec}$
Angular frequency, $\omega=2 \pi f=2 \pi \times 10^{12} / \mathrm{sec}$
Force constant, $k=m \omega^2=\frac{108 \times 10^{-3}}{602 \times 10^{23}} \times 4 \pi^2 \times 10^{24}$
$k=7.1 \mathrm{~N} / \mathrm{m}$
BETA
