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(A) Time-period in spring-block system is given by, $T = 2 \pi \sqrt{\frac{m}{k}} $If mass is doubled, $T^{\prime} = 2 \pi \sqrt{\frac{2m}{k}} $ $ \Ri

Vedclass · Accepted Answer

$ \begin{array}{llll} 3 & 2 & 1 & 4 \end{array} $

Vedclass · Answer

(A) Time-period in spring-block system is given by, $T = 2 \pi \sqrt{\frac{m}{k}} $If mass is doubled, $T^{\prime} = 2 \pi \sqrt{\frac{2m}{k}} $ $ \Rightarrow \quad T^{\prime} = \sqrt{2} T $Hence, time period becomes $ \sqrt{2} $ times, if the mass of the block is doubled.$ (A ightarrow 4) $(B) For spring, $ \mathrm{PE} = \frac{1}{2} k x^2 $If spring constant is increased 4 times, PE becomes 4 times.$ (B ightarrow 3) $(C) Speed of the block, $ v = \omega A $Hence, velocity of the block becomes twice when amplitude of the oscillation is doubled.$ (C ightarrow 2) $(D) Energy of the oscillation, $ E = \frac{1}{2} m \omega^2 a^2 $$ \begin{aligned} & & E^{\prime} = \frac{1}{2} m \omega^{\prime 2} a^2 \ \Rightarrow & E^{\prime} & = \frac{1}{2} m(2 \omega)^2 a^2 = \frac{4}{2} m \omega^2 a^2 \quad \left( \because \omega^{\prime} = 2 \omega ight) \ \Rightarrow & E^{\prime} & = 4 E \end{aligned} $Energy of the oscillation becomes 4 times when angular frequency is doubled.$ (D ightarrow 1) $

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

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