Answer:
Amplitude will be equal to 0.091 m
Explanation:
Given mass of the slits = 41 gram = 0.041 kg
Frequency f = 1.65 Hz
So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec[/tex]
Angular frequency is equal to [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]10.362 =\sqrt{\frac{k}{0.041}}[/tex]
Squaring both side
[tex]107.371 ={\frac{k}{0.041}}[/tex]
k = 4.40 N/m
For vertical osculation
[tex]mg=kA[/tex]
[tex]0.041\times 9.8=4.40\times A[/tex]
A = 0.091 m
So amplitude will be equal to 0.0391 m
