Two springs are connected to a block of mass M placed on a frictionless surface. If both the springs have a spring constant k, then the frequency of oscillation of the block is:
Show Hint
- $\frac{1}{2}\pi \sqrt{\frac{k}{M}}$
- $\frac{1}{2}\pi \sqrt{\frac{k}{2M}}$
- $\frac{1}{2}\pi \sqrt{\frac{2}{kM}}$
- $\frac{1}{2}\pi \sqrt{\frac{M}{k}}$
The Correct Option is C
Solution and Explanation
To find the frequency of oscillation for a block connected to two springs on a frictionless surface, let's consider the setup and apply relevant physics concepts.
The given system consists of two springs, each with spring constant \(k\), connected to a block of mass \(M\). The springs are arranged in parallel configuration, which means their equivalent spring constant can be added together. Thus, the equivalent spring constant \(k_{\text{eq}}\) is given by:
\(k_{\text{eq}} = k + k = 2k\)
The formula for the frequency of oscillation of a mass-spring system is:
\(f = \frac{1}{2\pi} \sqrt{\frac{k_{\text{eq}}}{M}}\)
Substitute the value of \(k_{\text{eq}}\) into the equation:
\(f = \frac{1}{2\pi} \sqrt{\frac{2k}{M}}\)
Hence, the frequency of oscillation is given by:
\(f = \frac{1}{2\pi} \sqrt{\frac{2k}{M}}\)
Therefore, the correct answer is:
\(\frac{1}{2\pi} \sqrt{\frac{2k}{M}}\)
This confirms the given correct answer option: \(\frac{1}{2\pi} \sqrt{\frac{2}{kM}}\), after simplifying the expression.
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