TS EAMCET · Physics · Oscillations
Two springs of spring constant \(k_1\) and \(k_2\) are connected by a mass \(m\) as shown in the figure. Under negligible friction, if the mass is displaced by small amount \(x\) from its equilibrium position and released, the period of oscillation is
- A \(2 \pi \sqrt{\frac{m\left(k_1+k_2\right)}{k_1 k_2}}\)
- B \(2 \pi \sqrt{\frac{m}{k_1+k_2}}\)
- C \(2 \pi \sqrt{\frac{m k_1 k_2}{\left(k_1+k_2\right)}}\)
- D \(2 \pi \sqrt{\frac{m\left(k_1-k_2\right)}{k_1 k_2}}\)
Answer & Solution
Correct Answer
(B) \(2 \pi \sqrt{\frac{m}{k_1+k_2}}\)
Step-by-step Solution
Detailed explanation
\( \text { According to question, } \) where, \(k_1=\) spring constant for first spring and \(k_2=\) spring constant for second spring. As, both the spring are in parallel connection, so spring constant equivalent \(k_p\) is \( k_p=k_1+k_2 \) The time period of oscillation for…
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