The natural frequency of a spring-mass system is 10 rad/s.
Which of the following statements is/are CORRECT?
Show Hint
- The mass is 100 kg and the stiffness is 1 N/m.
- The mass is 1.25 kg and the stiffness is 125 N/m.
- The stiffness is 620 N/m and the mass is 6.2 kg.
- The stiffness is 62 N/m and the mass is 620 kg.
The Correct Option is B
Solution and Explanation
Step 1: The natural frequency \( \omega_n \) of a spring-mass system is given by the formula: \[ \omega_n = \sqrt{\frac{k}{m}} \] where \( k \) is the stiffness of the spring (in N/m), and \( m \) is the mass (in kg).
Step 2: Given that the natural frequency \( \omega_n = 10 \, {rad/s} \), we can rearrange the formula to solve for the stiffness-mass relationship: \[ \omega_n^2 = \frac{k}{m} \Rightarrow k = m \omega_n^2 \] Substitute \( \omega_n = 10 \, {rad/s} \) into the equation: \[ k = m \times (10)^2 = 100m \] Step 3: Now, check each option to see which one satisfies this equation.
Option (A): If the mass is 100 kg and the stiffness is 1 N/m, then: \[ k = 100 \times 1 = 100 \, {N/m} \] But according to the formula, the stiffness should be \( k = 100 \times 100 = 10000 \, {N/m} \), so option (A) is incorrect.
Option (B): If the mass is 1.25 kg and the stiffness is 125 N/m, then: \[ k = 1.25 \times 100 = 125 \, {N/m} \] This satisfies the equation, so option (B) is correct.
Option (C): If the stiffness is 620 N/m and the mass is 6.2 kg, then: \[ k = 6.2 \times 100 = 620 \, {N/m} \] This also satisfies the equation, so option (C) is correct.
Option (D): If the stiffness is 62 N/m and the mass is 620 kg, then: \[ k = 620 \times 100 = 62000 \, {N/m} \] This does not satisfy the equation, so option (D) is incorrect.
Step 4: Therefore, the correct answers are options (B) and (C).
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