Question:
When a force \(F_1\) acts on a particle, frequency is 6 Hz and when a force \(F_2\) acts, frequency is 8 Hz. What is the frequency when both the forces act simultaneously in same direction?
When a force \(F_1\) acts on a particle, frequency is 6 Hz and when a force \(F_2\) acts, frequency is 8 Hz. What is the frequency when both the forces act simultaneously in same direction?
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For SHM, restoring force is proportional to displacement: \(F = -kx\). Frequency \(f \propto \sqrt{F_{\text{max}}}\).
Updated On: Apr 23, 2026
- 12 Hz
- 25 Hz
- 10 Hz
- 5 Hz
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For SHM, \(F \propto x\), so \(F = m\omega^2 x\). Also \(\omega = 2\pi f\). Thus \(F \propto f^2\).
Step 2: Detailed Explanation:
\(F_1 = k f_1^2\), \(F_2 = k f_2^2\). Total force \(F = F_1 + F_2 = k(f_1^2 + f_2^2)\).
But \(F = k f^2\). So \(f^2 = f_1^2 + f_2^2 = 6^2 + 8^2 = 36 + 64 = 100\).
\(f = 10\) Hz.
Step 3: Final Answer:
Thus, frequency = 10 Hz.
For SHM, \(F \propto x\), so \(F = m\omega^2 x\). Also \(\omega = 2\pi f\). Thus \(F \propto f^2\).
Step 2: Detailed Explanation:
\(F_1 = k f_1^2\), \(F_2 = k f_2^2\). Total force \(F = F_1 + F_2 = k(f_1^2 + f_2^2)\).
But \(F = k f^2\). So \(f^2 = f_1^2 + f_2^2 = 6^2 + 8^2 = 36 + 64 = 100\).
\(f = 10\) Hz.
Step 3: Final Answer:
Thus, frequency = 10 Hz.
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