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Standard XII

Physics

Spring Block's Time Period

The right bod...

Question

The right bodies $A$ and $B$ of masses $2 kg$ and $4 kg$ respectively are rigidly connected to a spring of the constant $400 N/m$ . The body $B$ rests on a horizontal table, from the rest position, the body $A$ is compressed by $2 cm$ and then released. Find the frequency of oscillation.

1.60 Hz

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2.25 Hz

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1.12 Hz

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3.18 Hz

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Solution

The correct option is B $2.25 Hz$
Given,
Mass of body $A (m_{A}) = 2 kg$
Mass of body $B (m_{B}) = 4 kg$
Force constant $(K) = 400 N/m$
Frequency of oscillation $(v) = \frac{1}{2 π} \sqrt{\frac{K}{m_{A}}}$
$= \frac{1}{2 π} \sqrt{\frac{400}{2}}$
$= \frac{20}{2 \sqrt{2} π} = 2.25 Hz$

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Spring Block's Time Period

Standard XII Physics

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The right bodies A and B of masses 2 kg and 4 kg respectively are rigidly connected to a spring of the constant 400 N/m. The body B rests on a horizontal table, from the rest position, the body A is compressed by 2 cm and then released. Find the frequency of oscillation.

The correct option is B 2.25 HzGiven, Mass of body A (mA)=2 kg Mass of body B (mB)=4 kg Force constant (K)=400 N/m Frequency of oscillation (v)=12π√KmA =12π√4002 =202√2π=2.25 Hz

The right bodies $A$ and $B$ of masses $2 kg$ and $4 kg$ respectively are rigidly connected to a spring of the constant $400 N/m$ . The body $B$ rests on a horizontal table, from the rest position, the body $A$ is compressed by $2 cm$ and then released. Find the frequency of oscillation.

The correct option is B $2.25 Hz$
Given,
Mass of body $A (m_{A}) = 2 kg$
Mass of body $B (m_{B}) = 4 kg$
Force constant $(K) = 400 N/m$
Frequency of oscillation $(v) = \frac{1}{2 π} \sqrt{\frac{K}{m_{A}}}$
$= \frac{1}{2 π} \sqrt{\frac{400}{2}}$
$= \frac{20}{2 \sqrt{2} π} = 2.25 Hz$