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  • When a mass m is connected individually to two springs S_1 and S_2, the oscillation frequencies are ν_1 and ν_2. If the same mass is attached to the two springs, as shown in Figure , the oscillation frequency would be

When a mass m is connected individually to two 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 Figure, the oscillation frequency would be



 

Answers (1)

Answer: The answer is the option (b) \sqrt{v_{1}^{2}+v_{2}^{2}}

Explanation:  If we connect a mass (m) to a spring on a frictionless horizontal surface, then, their frequencies will be –

V_{1} = \frac{1}{2\pi}\sqrt{\frac{K_{1}}{m}}    &   V_{2} = \frac{1}{2\pi}\sqrt{\frac{K_{2}}{m}}

   …. (i)                   …. (ii)

Since the springs are parallel, their equivalent will be-

K_{p} = K_{1} + K_{2}

& frequency will be –

V_{p} = V_{1} =\frac{ 1}{2 \pi}\sqrt{\frac{K_{1} + K_{2}}{m}}

  =\frac{ 1}{2 \pi}\left [ \frac{K_{1}}{m}+\frac{K_{2}}{m} \right ]^{\frac{1}{2}}

From (i),

\frac{ K_{1}}{m} = (2\pi v_{1})^{2} = 4\pi^{ 2}v_{1}^{2}

& \frac{ K_{2}}{m} = (2\pi v_{1})^{2} = 4\pi^{ 2}v_{2}^{2}

Thus, V_{p} = \frac{1}{2\pi} [4\pi ^{2}v_{1}^{2} + 4\pi ^{2}v_{2}^{2}]

V_{p} = \sqrt{v_{1}^{2}+v_{2}^{2}}

Hence, opt (b)

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