91影视

Following are the data given in the question,

Mass$m=300g$

Displacement$x_1=3.0cm$

Speed$v_1=95.4cm/s$

Displacement$x_2=6.0cm$

Speed$v_1=71.4cm/s$.

The displacement of oscillator is $x\left(t\right)=A\cos 鈦�(蝇t+蠒)$

The Velocity of oscillator is $v\left(t\right)=鈭�v_{max}\sin 鈦�(蝇t+蠒)$

where, amplitude and maximum velocity can be expressed as A = $\frac{v_{max}}{蝇}$

In the given question, we have two moments, this can be expressed as

Displacement $x_1\left(t\right)=\frac{v_{max}}{蝇}\cos 鈦�\left(蝇t_1\right)$--- (a1)

Displacement $x_2\left(t\right)=\frac{V_{max}}{蝇}\cos \left(蝇t_2\right)$---(a2)

Velocity $v_1\left(t\right)=-v_{max}\sin \left(蝇t_1\right)$--- (b1)

Velocity $v_2\left(t\right)=-V_{max}\sin \left(蝇t_2\right)$---(b2)

where$蝇$represents angular frequency

we can get $v_{max}$by squaring (a1), (a2), (b1) and (b2)

we can use the know expression, ${\sin }^2\left(t\right)+{\cos }^2\left(t\right)=1$

$v_{max}^2=v_1^2+x_1^2鈰�{蝇}^2$

$v_{max}^2=v_2^2+x_2^2鈰�{蝇}^2$.

Substituting the given values in the expression,

$v_{max}^2=\frac{v_1^2鈭�v_2^2鈰�\frac{x_1^2}{x_2^2}}{1鈭�\frac{x_1^2}{x_2^2}}$

$v_{max}^2=10435.56$

Therefore, the maximum velocity is$102.1$cm/s.