When the length I of a piece of wire under constant tension is varied, the relationship of the frequency of the vibration f\(\alpha\)/ is
a
f\(\alpha\)/
b
f\(\alpha \frac{1}{I^2}\)
c
f \(\alpha\sqrt I\)
d
f\(\alpha \frac{1}{1}\)
Explanation
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VPETERS2002
6 years ago
The period of motion T of the simple pendulum of length L can be shown to equal to
T=2pi*square root of (L/g)
where 2pi is a constant and g(acceleration due to gravity) remains unchanged. The above expression is valid provided that the angular displacement or the the amplitude of motion is very small
Therefore, T is directly proportional to square root of(L)
Now;
T=k*square root of (L)
T=1/F
1/F=k*square root of (L)
cross multiply
F * k*square root of (L)=1
make F the subject of the formula
F=1/k*square root of (L)
since k is a constant
F is directly proportional to 1/square root of (L)......ANSWER
