Two identical blocks P and Q have mass ‘m’ each. They are attached to two identical springs initially unstretched. Now the left spring (along with P) is compressed by $\frac{A}{2}$ and the right spring (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initial time period of both the blocks was T. [consider k as spring constant of each spring]

The time period and amplitude of oscillation of the combined mass is respectively,

$\frac{T}{\sqrt{2}}$ and $\frac{A}{4}$

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$\frac{T}{2}$ and $\frac{A}{2}$

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$T$ and $\frac{A}{4}$

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$\frac{T}{2}$ and $\frac{A}{4}$

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Solution

The correct option is C
$T$ and $\frac{A}{4}$

Time Period:
They will collide at their mean positions because time period of both are same and that is $2 π \sqrt{\frac{m}{k}}$ . After collision combined mass is 2m and $k_{e f f} = 2 k$ . Hence, time period remains unchanged.
Amplitude:
Applying conservation of linear momentum, velocity of combined mass just after collision is $v = \frac{A ω}{4}$

Since this is the velocity at mean position,
Hence, $v = ω^{^{'}} A^{^{'}}$
or $\frac{ω A}{4} = ω^{^{'}} A^{^{'}}$
or $A^{^{'}} = \frac{A}{4} a s ω^{^{'}} = ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{2 k}{2 m}}$

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Two identical blocks P and Q have mass ‘m’ each. They are attached to two identical springs initially unstretched. Now the left spring (along with P) is compressed by $\frac{A}{2}$ and the right spring (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initial time period of both the blocks was T. [consider K as spring constant of each spring]