Answer:
The pendulum should be made longer by 0.194m in order to increase its period by 0.32s
Explanation:
using the formula T= 2π[tex]\sqrt{\frac{L}{g} }[/tex]
rearranging the equation and making L subject of formula we have;
L=T²g/4π²
lets calculate the length when T=1.06s
g=9.8m/s² , π=3.124
[tex]L=\frac{1.06^{2}*9.8 }{4*3.142^{2} }[/tex]
L=0.279m
the new period after its increased by 0.32s = 1.06+0.32 =1.38s
[tex]L_{2}=\frac{1.38^{2}*9.8 }{4 *3.142^{2} }[/tex]
[tex]L_{2}=0.473m[/tex]
increase in length = 0.473 -0.279
=0.194m
The pendulum should be "0.194 m" longer.
According to the question,
Time,
We know,
By using the formula,
→ [tex]T = 2 \pi \sqrt{\frac{L}{g} }[/tex]
or,
→ [tex]L = \frac{T^2g}{4 \pi^2}[/tex]
By substituting the values, we get
→ [tex]= \frac{1.06^2\times 9.8}{4\times 3.142^2}[/tex]
→ [tex]= 0.279 \ m[/tex]
Now,
The new period after it increased by 0.32 s, we get
= [tex]1.06+0.32[/tex]
= [tex]1.38 \ s[/tex]
then,
→ [tex]L_2 = \frac{1.38^2\times 9.8}{4\times 3.142^2}[/tex]
[tex]= 0.473 \ m[/tex]
hence,
The increase in length will be:
= [tex]L_2-L[/tex]
= [tex]0.473-0.279[/tex]
= [tex]0.194 \ m[/tex]
Thus the answer above is right.
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