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