AIIMS2003Physics-Oscillations

AIIMS 2003 Physics Spring Mass System MCQ Question

Type: MCQ-conceptual-Medium-Class 11

Two springs of force constants k and 2k are connected to a mass as shown in figure. The frequency of oscillation of the mass is

Question diagram
A

1/2π √(k/m)

B

1/2π √(2k/m)

C

1/2π √(3k/m)

D

1/2π √(k/m)

Correct Answer

Option C

Detailed Explanation

To determine the frequency of oscillation of a mass connected to two springs with force constants kk and 2k2k, we first need to find the equivalent spring constant when the springs are arranged in parallel.

Step 1: Understanding the Configuration

Assuming the two springs are connected in parallel to the mass, the equivalent spring constant KK can be calculated as:

K=k1+k2K = k_1 + k_2

where k1=kk_1 = k (the spring constant of the first spring) and k2=2kk_2 = 2k (the spring constant of the second spring). Therefore, we have:

K=k+2k=3kK = k + 2k = 3k

Step 2: Frequency of Oscillation

The frequency of oscillation ff of a mass-spring system is given by the formula:

f=12πKmf = \frac{1}{2\pi} \sqrt{\frac{K}{m}}

where KK is the equivalent spring constant and mm is the mass attached to the springs. Substituting the equivalent spring constant we found:

f=12π3kmf = \frac{1}{2\pi} \sqrt{\frac{3k}{m}}

Conclusion

Thus, the frequency of oscillation of the mass is:

f=12π3kmf = \frac{1}{2\pi} \sqrt{\frac{3k}{m}}

This matches with option C, confirming that the correct answer is C) 12π3km\frac{1}{2\pi} \sqrt{\frac{3k}{m}}.

Explanation of Incorrect Options

Now, let's clarify why the other options are incorrect:

  • Option A: 12πkm\frac{1}{2\pi} \sqrt{\frac{k}{m}}

    This option would be valid if there were only one spring with a spring constant kk. However, since we have two springs with constants kk and 2k2k, this does not account for the combined effect of both springs.

  • Option B: 12π2km\frac{1}{2\pi} \sqrt{\frac{2k}{m}}

    This would be true if the equivalent spring constant was 2k2k, which again neglects the contribution of the spring kk. Thus, it only considers one spring.

  • Option D: 12πkm\frac{1}{2\pi} \sqrt{\frac{k}{m}}

    Similar to option A, this option only considers the spring with spring constant kk and ignores the second spring, leading to an incorrect calculation.

Summary

To summarize, the calculated equivalent spring constant for the two springs in parallel is 3k3k. Therefore, the frequency of oscillation of the mass is given by:

f=12π3kmf = \frac{1}{2\pi} \sqrt{\frac{3k}{m}}

This confirms that the correct answer is option C) 12π3km\frac{1}{2\pi} \sqrt{\frac{3k}{m}}.

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