Chapter 15: 41 - Excercises And Problems (page 416)

A $0.300 k g$ oscillator has a speed of $95.4 c m / s$ when its displacement is role="math" localid="1651383318687" $3.00 c m$ and $71.4 c m / s$ when its displacement is $6.00 c m$ . What is the oscillator’s maximum speed?

Short Answer

Expert verified

Using the information provided in the problem, the oscillator's maximum speed is $102.1 cm / s .$

Step by step solution

Step: 1 Equating:

The oscillator displacement is

$x (t) = A \cos (ω t + ϕ)$

The velocity oscillation is

$v (t) = - v_{max} \sin (ω t + ϕ)$

The maximum velocity and amplitude as

$A = \frac{v_{max}}{ω}$

We have two moments as

Displacement at one , $x_{1} (t) = \frac{v_{max}}{ω} \cos (ω t_{1})$

Displacement at two, $x_{2} (t) = \frac{V_{m a x}}{ω} \cos (ω t_{2})$

Step: 2 Calculation:

Velocity at $v_{1} (t) = - v_{m a x} \sin (ω t_{1})$ and

Velocity at $v_{2} (t) = - V_{m a x} \sin (ω t_{2})$

where $ω$ is angular frequency.

By squaring above equation we get $v_{m a x}$ .

we can use expression as $\sin^{2} (t) + \cos^{2} (t) = 1$ ,we get

role="math" localid="1651384339020" $v_{max}^{2} = v_{1}^{2} + x_{1}^{2} \cdot ω^{2}$

$v_{max}^{2} = v_{2}^{2} + x_{2}^{2} \cdot ω^{2}$

Step: 3 Solution:

Substituting values in above expression,we get as

$v_{max}^{2} = \frac{v_{1}^{2} - v_{2}^{2} \times \frac{x_{1}^{2}}{x_{2}^{2}}}{1 - \frac{x_{1}^{2}}{x_{2}^{2}}}$

$v_{max}^{2} = 10435.56$

Therefore, the maximum velocity is $v_{max}^{2} = 102.1 m / s$ .

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A $0.300 k g$ oscillator has a speed of $95.4 c m / s$ when its displacement is role="math" localid="1651383318687" $3.00 c m$ and $71.4 c m / s$ when its displacement is $6.00 c m$ . What is the oscillator’s maximum speed?

Short Answer

Step by step solution

Step: 1 Equating:

Step: 2 Calculation:

Step: 3 Solution:

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A 0.300kg oscillator has a speed of 95.4cm/s when its displacement is role="math" localid="1651383318687" 3.00cm and 71.4cm/s when its displacement is 6.00cm. What is the oscillator’s maximum speed?

Short Answer

Step by step solution

Step: 1 Equating:

Step: 2 Calculation:

Step: 3 Solution:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Waves Physics

Turning Points in Physics

Rotational Dynamics

Magnetism

Physics of Motion

Famous Physicists

Study anywhere. Anytime. Across all devices.

A $0.300 k g$ oscillator has a speed of $95.4 c m / s$ when its displacement is role="math" localid="1651383318687" $3.00 c m$ and $71.4 c m / s$ when its displacement is $6.00 c m$ . What is the oscillator’s maximum speed?