Question

A 0.300 kg oscillator has a speed of 95.4 cm/s when its displacement is 3.00 cm and 71.4 cm/s when its displacement is 6.00 cm. What is the oscillator’s maximum speed?

Short answer

Using the data given in the question, the maximum speed of the oscillator is $102.1\mathrm{cm}/\mathrm{s}$.

Step-by-step solution

Introduction

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$v_1=71.4cm/s$.

Calculation

The displacement of oscillator is $x\left(t\right)=A\cos (\omega t+\varphi )$

The Velocity of oscillator is $v\left(t\right)=-v_{max}\sin (\omega t+\varphi )$

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

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

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

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

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

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

where$\omega$represents angular frequency

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

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

$v_{max}^2=v_1^2+x_1^2\cdot {\omega }^2$

$v_{max}^2=v_2^2+x_2^2\cdot {\omega }^2$.

Final Solution

Substituting the given values in the expression,

$v_{max}^2=\frac{v_1^2-v_2^2\cdot \frac{x_1^2}{x_2^2}}{1-\frac{x_1^2}{x_2^2}}$

$v_{max}^2=10435.56$

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