AP EAMCET · PHYSICS · Oscillations
When a mass ' \(m\) ' is connected individually to the springs \(s_1\) and \(s_2\), the oscillation frequencies are \(v_1\) and \(v_2\). If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be
- A \(v_1+v_2\)
- B \(\sqrt{v_1^2+v_2^2}\)
- C \(\left(\frac{1}{v_1}+\frac{1}{v_2}\right)^{-1}\)
- D \(\sqrt{v_1^2-v_2^2}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{v_1^2+v_2^2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \delta_1=\sqrt{\frac{\mathrm{S}_1}{\mathrm{~m}}}, \delta_2=\sqrt{\frac{\mathrm{S}_2}{\mathrm{~m}}} \\ & \delta=\sqrt{\frac{\mathrm{S}_{\mathrm{eq}}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{S}_1+\mathrm{S}_2}{\mathrm{~m}}}=\sqrt{\frac{\mathrm{m}…
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