Answer:
[tex]\huge\boxed{\sf f=0.55 \ Hz}[/tex]
Explanation:
Given Data:
Length = l = 820 mm = 0.82 m
Acceleration due to gravity = g = 9.8 ms⁻²
Required:
Frequency = f = ?
Formula:
[tex]\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }[/tex]
Solution:
[tex]\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}[/tex]
