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Byju's Answer

Standard XII

Physics

Spring Block's Time Period

A 1 kg mass...

Question

A $1 k g$ mass is attached to a spring of force constant $600 N / m$ and rests on a smooth horizontal surface with other end of the spring tied to wall as shown in figure. A second mass of $0.5 k g$ slides along the surface toward the first at $3 m / s$ . If the masses make a perfectly inelastic collision, then find amplitude and time period of oscillation of combined mass.

5 c m, \frac{π}{10} s

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5 c m, \frac{π}{5} s

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4 c m, \frac{2 π}{5} s

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4 c m, \frac{π}{3} s

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Solution

The correct option is A $5 c m, \frac{π}{10} s$
Applying linear momentum conservation,
$0.5 \times 3 = (1 + 0.5) v$ or $v = 1 m / s$
By conversation of energy,
After collision
$\frac{1}{2} (1 + 0.5) v^{2} = \frac{1}{2} k A^{2} \Rightarrow A = \sqrt{\frac{1.5}{k}} \times v$
$\Rightarrow A = \sqrt{\frac{1.5}{600}} \times 1 = \frac{1}{20} m = 0.05 m$
$A = 5 c m$ .
Time period of oscillation,
$T = 2 π \sqrt{\frac{m_{1} + m_{2}}{k}} = 2 π \sqrt{\frac{1.5}{600}} = \frac{2 π}{20} = \frac{π}{10} s$ .

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The correct option is A 5 cm,π10sApplying linear momentum conservation,0.5×3=(1+0.5)v or v=1m/sBy conversation of energy,After collision12(1+0.5)v2=12kA2⇒A=√1.5k×v⇒A=√1.5600×1=120m=0.05 mA=5 cm.Time period of oscillation,T=2π√m1+m2k=2π√1.5600=2π20=π10s.