Answer:83.17 cm/s
Explanation:
Let positive x be negative direction and negative x be positive direction in this question
General equation of motion of SHM is
x=[tex]Asin\left ( \omega_{n}t\right )[/tex] --------1
where [tex]\omega [/tex]is natural frequency of motion given by
[tex]\omega_n[/tex]=[tex]\sqrt{\frac{k}{m}}[/tex]
Where K is spring constant
here A=23.7cm
And it is given it is again at 23.7 from equilibrium position having passed through the equilibrium once.
i.e. it covers this distance in [tex]\frac{T}{2}[/tex] sec
where T is the time period of oscillation i.e. returning to same place after T sec
therefore T=0.634 sec
differentiating equation 1 we get
v=[tex]A\omega_n[/tex]cos[tex]\left ( \omega_{n}t\right )[/tex]
and [tex]T\times \omega_n[/tex]=[tex]2\pi[/tex]
[tex]\omega_n[/tex]=9.911rad/s
[tex]v[/tex]=[tex]23.7\times 9.911cos\left (789.261\degree\right)[/tex]
v=83.17 cm/s
