Answer:
1.57 kg
Explanation:
Let the mass of block is m.
Spring constant, K = 9.4 N/m
Amplitude, A = 15.5 cm
y = A/2
v = 32.8 cm/s = 0.328 m/s
The velocity of the particle executing SHM is given by
[tex]v=\omega\sqrt{A^{2}-y^{2}}[/tex]
where, ω is the angular frequency of SHM.
[tex]0.328=\omega\sqrt{A^{2}-\left ( \frac{A}{2} \right )^{2}}[/tex]
[tex]0.328=\omega\times 0.866 A[/tex]
0.328 = ω x 0.866 x 0.155
ω = 2.45 rad/s
Now the angular frequency is given by
[tex]\omega = \sqrt\frac{K}{m}[/tex]
K = mω²
9.4 = m x 2.45 x 2.45
m = 1.57 kg