Ncert Exercises & Solutions

When a mass $m$ is connected individually to two springs $S_1$ and $S_2,$ the oscillation frequencies are $\nu_1$ and $\nu_2.$ If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be:

1.	$v_2+v_2$	2.	$\sqrt{v_1^2+v_2^2}$
3.	$\left(\dfrac{1}{v_1}+\dfrac{1}{v_1}\right)^{-1}$	4.	$\sqrt{v_1^2-v_2^2}$

Hint:

ω = 2 π ν = \sqrt{\frac{k}{m}}

$\omega = 2\pi \nu = \sqrt{\frac{k}{m}}$

Step 1: Find the frequency of the spring-mass system.

The angular frequency

ω

$\omega$ of a mass-spring system is given by;

ω = 2 π ν = \sqrt{\frac{k}{m}}

$\omega = 2\pi \nu = \sqrt{\frac{k}{m}}$
For each spring, we have;

ν_{1} = \frac{1}{2 π} \sqrt{\frac{k_{1}}{m}}, ν_{2} = \frac{1}{2 π} \sqrt{\frac{k_{2}}{m}}

$\nu_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}, \quad \nu_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}}$
where

k_{1}

$k_1$ and

k_{2}

$k_2$ are the spring constants of springs

S_{1}

$S_1$ and

S_{2}

$S_2$ respectively.

Step 2: Find the effective spring constant $k_{\text{eff}}.$

When two springs are connected in parallel, their effective spring constant is given by;
$\Rightarrow k_{\text{eff}} = k_1 + k_2$

Step 3: Find the individual frequencies.

When the mass is connected to the springs individually:

ν_{1} = \frac{1}{2 π} \sqrt{\frac{k_{1}}{m}}, ν_{2} = \frac{1}{2 π} \sqrt{\frac{k_{2}}{m}}

$\nu_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}, \quad \nu_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}}$
Squaring both sides we get;

ν_{1}^{2} = \frac{k_{1}}{4 π^{2} m} \Rightarrow k_{1} = 4 π^{2} m ν_{1}^{2}

$\nu_1^2 = \frac{k_1}{4\pi^2 m}\Rightarrow k_1= 4\pi^2m\nu^2_1$

ν_{2}^{2} = \frac{k_{2}}{4 π^{2} m} \Rightarrow k_{2} = 4 π^{2} m ν_{2}^{2}

$\nu_2^2 = \frac{k_2}{4\pi^2 m}\Rightarrow k_2 = 4\pi^2m\nu^2_2$

\Rightarrow k_{eff} = k_{1} + k_{2}

$\Rightarrow k_\text{eff} = k_1 +k_2$

$\Rightarrow \nu_\text{eff}^2=\nu_1^2+\nu_2^2$
Taking the square root we get;
$\Rightarrow \nu_{\text{eff}} = \sqrt{\nu_1^2 + \nu_2^2}$
Hence, option (2) is the correct answer.

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