Answer:
The new time period is  [tex]T_2 = 3.8 \ s[/tex]
Explanation:
From the question we are told that
 The period of oscillation is  [tex]T = 5 \ s[/tex]
  The  new  length is  [tex]l_2 = 0.76 \ m[/tex]
Let assume the original length was [tex]l_1 = 1 m[/tex]
Generally the time period is mathematically represented as
     [tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]
Now  I is the moment of inertia of the stick which is mathematically represented as
      [tex]I = \frac{m * l^2 }{12 }[/tex]
So
    [tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]
Looking at the above equation we see that
    [tex]T \ \ \ \alpha \ \ \ l[/tex]
=> Â Â [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]
=> Â Â [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]
=> Â Â [tex]T_2 = 3.8 \ s[/tex]
