Question
Download Solution PDF200 kg द्रव्यमान, 80 N/mm कठोरता वाले स्प्रिंग और 800 N-s/m अवमंदन गुणांक वाले अवमंदक से युक्त एक कंपन प्रणाली है। अवमंदित कंपन की प्राकृतिक आवृत्ति क्या होगी?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंप्रत्यय:
एक कंपन प्रणाली की अवमंदित प्राकृतिक आवृत्ति निम्न द्वारा दी जाती है
\( \omega_d = \omega_n \sqrt{1 - \zeta^2} \), जहाँ \( \omega_n = \sqrt{\frac{k}{m}} \) प्राकृतिक आवृत्ति है और \( \zeta = \frac{c}{2\sqrt{km}} \) अवमंदन अनुपात है।
दिया गया है:
द्रव्यमान, m = 200 kg
स्प्रिंग कठोरता, k = 80 N/mm = 80000 N/m
अवमंदन गुणांक, c = 800 Ns/m
परिकलन:
अनावमंदित प्राकृतिक आवृत्ति, \( \omega_n = \sqrt{\frac{80000}{200}} = \sqrt{400} = 20~\text{rad/s} \)
अवमंदन अनुपात, \( \zeta = \frac{800}{2\sqrt{80000 \cdot 200}} = \frac{800}{2 \cdot 4000} = 0.1 \)
अवमंदित प्राकृतिक आवृत्ति, \( \omega_d = 20 \cdot \sqrt{1 - 0.1^2} = 20 \cdot \sqrt{0.99} = 19.8998~\text{rad/s} \)
Hz में अवमंदित आवृत्ति, \( f_d = \frac{\omega_d}{2\pi} = \frac{19.8998}{6.283} \approx 3.17~\text{Hz} \)
अब, \( \frac{3\sqrt{11}}{\pi} = \frac{3 \times 3.3166}{3.14} \approx \frac{9.9498}{3.14} \approx 3.17~\text{Hz} \)
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