Answer:
f= 1.75 Hz
Explanation:
Given that
Spring constant ,K = 179 N/m
Mass of the ball ,m = 1.48 kg
Lets take angular frequency = ω rad/s
We know that
ω² = m K
[tex]\omega=\sqrt{\dfrac{K}{m}}[/tex]
Now by putting the values in the above equation
[tex]\omega=\sqrt{\dfrac{179}{1.48}}[/tex]
ω= 10.99 rad/s
We know that
ω = 2 π f
[tex]f=\dfrac{\omega}{2\pi }[/tex]
[tex]f=\dfrac{10.99}{2 \pi }[/tex]
f= 1.75 Hz
Therefore the frequency will be 1.75 Hz.
