Correct Answer is Option (A)

F = Kx

10 = K $\times$ 0.05

K = $\frac{1000}{5}=200$

$T=2\pi \sqrt{\frac{m}{K}}=2\pi \sqrt{\frac{2}{200}}$

$=\frac{2\pi }{10}=\frac{6.28}{10}$

= .628 s

Correct Answer is Option (B)

Let us assume, the length of spring be l.

When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths $\frac{l}{6},\frac{2l}{6}$ and $\frac{3l}{6}$ with force constant,

$\therefore$ $k_1=\frac{kl}{l_1}=\frac{kl}{l/6}=6k$

$k_2=\frac{kl}{l_2}=\frac{kl}{2l/6}=3k$

$k_3=\frac{kl}{l_3}=\frac{kl}{3l/6}=2k$

When connected in series,

$\frac{1}{k^{\prime }}=\frac{1}{6k}+\frac{1}{3k}+\frac{1}{2k}=\frac{1+2+3}{6k}=\frac{1}{k}$

$\therefore k^{\prime }=k$

When connected in parallel,

k'' = 6k + 3k + 2k = 11k

$\frac{k^{\prime }}{k^{″}}=\frac{k}{11k}=\frac{1}{11}$

Correct Answer is Option (D)

The time period of oscillation is given by

$T=2\pi \sqrt{m/k}$

$T_1=3=2\pi \sqrt{m/k}$

$T_2=5=2\pi \sqrt{\left(\frac{m+1}{k}\right)}$

On dividing :

$3/5=\sqrt{m/\left(m+1\right)}$

$9/25=\sqrt{m/\left(m+1\right)}$

$9m+9=25m$

$16m=9\Rightarrow m=9/16$

Correct Answer is Option (D)

$T=2\pi \sqrt{\frac{m}{K}}$

$\therefore \frac{T_1}{T_2}=\sqrt{\frac{M_1}{M_2}}$

$\therefore$ $T_2=T_1\sqrt{\frac{M_2}{M_1}}=T_1\sqrt{\frac{2M}{M}}$

$T_2=T_1\sqrt{2}=\sqrt{2}T$     (where T₁ =T)

Correct Answer is Option (A)

class="MJX-TEX" aria-hidden="true">//-->$\Rightarrow a=\frac{2\times 10}{200}$   $[\because As={\omega }^2=\frac{k}{m}]$

$\Rightarrow$ a = 1/10 m = 10 cm

Correct Answer is Option (D)

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\Rightarrow \frac{1}{k_{eq}}=\frac{k_1+k_2}{k_1{k}_2}$

$\Rightarrow k_{eq}=\frac{k_1{k}_2}{k_1+k_2}$

Correct Answer is Option (C)

Let k be the force constant of spring. If k' is the force constant of each part, then

$\frac{1}{l}=\frac{4}{k^{\prime }}\Rightarrow k^{\prime }=4k$

$\therefore$ Time period
= $2\pi \sqrt{\frac{m}{4k}}=\frac{1}{2}\times 2\pi \sqrt{\frac{m}{k}}=\frac{T}{2}$

Correct Answer is Option (B)

The time period of a spring mass system as shown in figure 1 is given by $T=2\pi \sqrt{m/k}$, where k is the spring constant.

$\therefore$ $t_1=2\pi \sqrt{m/k_1}$   ...(i)
and $t_2=2\pi \sqrt{m/k_2}$   ...(ii)


Now, when they are connected in parallel as shown in figure 2(a), the system can be replaced by a single spring of spring constant, k_eff = k₁ + k₂.
[Since mg = k₁x + k₂x = k_effx]

$\therefore$ $t_0=2\pi \sqrt{m/k_{eff}}=2\pi \sqrt{m/\left(k_1+k_2\right)}$   ...(iii)

From (i), $\frac{1}{t_1^2}=\frac{1}{4{\pi }^2}\times \frac{k_1}{m}$   ...(iv)

From (ii), $\frac{1}{t_2^2}=\frac{1}{4{\pi }^2}\times \frac{k_2}{m}$   ...(v)

From (ii), $\frac{1}{t_0^2}=\frac{1}{4{\pi }^2}\times \frac{k_1+k_2}{m}$   ...(vi)

Now (iv) + (v)

$\frac{1}{t_1^2}+\frac{1}{t_2^2}=\frac{1}{4{\pi }^{2}m}\left(k_1+k_2\right)=\frac{1}{t_0^2}$

$\therefore$ $t_0^{-2}=t_1^{-2}+t_2^{-2}$