Question

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 1.9 m/s. Determine the amplitude A of the motion. A = Entry field with correct answer .07756

Answer 1

Answer:

The amplitude of the motion is 0.077 m

Explanation:

It is given that,

Mass of the object, m = 0.2 kg

Spring constant, K = 120 N/m

Maximum speed of the object, v = 1.9 m/s

We know that, in harmonic motion, the maximum speed of the object is given by :

$v_{max}=A\omega$

Where

A is the amplitude of the wave

$\omega$ is the angular velocity, $\omega=\sqrt{\dfrac{k}{m}}$

$v_{max}=A\sqrt{\dfrac{k}{m}}$

$A=v_{max}\sqrt{\dfrac{m}{K}}$

$A=1.9\times \sqrt{\dfrac{0.2}{120}}$

A = 0.077 m

So, the amplitude of the motion is 0.077 m. Hence, this is the required solution.                                          

Answer 2

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$  where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$