Given situation is as shown below.
Speed of approach of source
$=v_s=r \omega=30 \,ms ^{-1}$
Using formula for Doppler's effect,
Maximum frequency,
$f_{\max }=\left(\frac{v}{v-v_s}\right) \cdot f$
Minimum frequency,
$f_{\min }=\left(\frac{v}{v+v_s}\right) \cdot f$
So, ratio of
$\frac{f_{\max }}{f_{\min }}=\frac{v+v_s}{v-v_s}=\frac{330+30 }{330-30}$
$\Rightarrow f_{\max }=1.2$
$(A)$ $u=0.8 v$ and $f_5=f_0$
$(B)$ $u=0.8 v$ and $f_5=2 f_0$
$(C)$ $u=0.8 v$ and $f_5=0.5 f_0$
$(D)$ $u=0.5 v$ and $f_5=1.5 f_0$
$y = 0.02sin\left[ {2\pi \left( {\frac{t}{{0.04\left( s \right)}} - \frac{x}{{0.50\left( m \right)}}} \right)} \right]m$ The tension in the string is .... $N$