Given, amplitude of spring, $A=50 \times 10^{-2} \mathrm{~m}$, maximum frequency heard by observer, $n_{\text {max }}=1.125 n_{\text {min }}$ mass of sound source, $m=100 \mathrm{~g}=0.10 \mathrm{~kg}$ and speed of sound, $v=340 \mathrm{~ms}^{-1}$.

As, the apparent frequency heard by observer for moving source,
$$
n=\frac{n_0 v}{v-v_s}
$$

Hence, for $n_{\max }, v_s=v_{s \max }$
and for $\quad n_{\min }, v_s=-v_{s \max }$


By dividing Eqs. (i) by (ii), we get
$$
\begin{aligned}
& \frac{1.125 n_{\min }}{n_{\min }}=\frac{v+v_{s \max }}{v-v_{s \max }} \\
& \Rightarrow \quad 1.125\left(v-v_{s \max }\right)=\left(v+v_{s \max }\right) \\
& \Rightarrow \quad v(0.125)=2.125 v_{s \max } \\
& \Rightarrow \quad \frac{340 \times 0.125}{2.125}=v_{s \max } \\
&
\end{aligned}
$$

From Eqs. (iii) and (iv), we get
$$
\begin{aligned}
k & =\frac{v_{s \max }^2 m}{A^2} \\
\Rightarrow \quad k & =\frac{20 \times 20 \times 0.10}{0.50 \times 0.50}=160 \mathrm{~N} \mathrm{~m}^{-1}
\end{aligned}
$$

Hence, the correct option is (c).