JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2017
- Medium
- MCQ
A \(1 \,kg\) block attached to a spring vibrates with a frequency of \(1\, Hz\) on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an \(8\, kg\) block placed on the same table. So, the frequency of vibration of the \(8\, kg\) block is ..... \(Hz\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(0.5\)
Step-by-step Solution
Detailed explanation
Frequency of spring \((f)=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=1 \mathrm{Hz}\)
\(\Rightarrow 4 \pi^{2}=\frac{\mathrm{k}}{\mathrm{m}}\)
If block of mass \(\mathrm{m}=1 \mathrm{kg}\) is attached then, \(k=4 \pi^{2}\)
Now, identical springs are attached in parallel with mass
\(\mathrm{m}=8 \mathrm{kg}\), Hence, \(k_{e q}=2 k\)
\(f^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k} \times 2}{8}}=\frac{1}{2} \mathrm{Hz}\)
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