Question

two point masses 3kg and 1 kg are attached to opposite ends of a horizontal spring whose spring constant is 300 N/m the natural vibration frequency is
​

Answer 1

Answer:

Explanation:

k = 300 n/m

mass are in opposite direction so 3-1= 2

w= under root of 300/2 = under root 150 = 12.247

hope that's the answer to the question

thanks

Answer 2

The natural vibration frequency is  3 Hz.

​Explanation:

The natural frequency of vibration of the two points masses system is

given as

$\nu=\frac{1}{2\pi} \sqrt{\frac{k}{\mu} }$

Here $\mu$ is the reduce mass of the system and k is the spring constant.

Reduce mass,

$\mu=\frac{m_1m_2}{m_1+m_2}$

$\mu=\frac{3kg\times1kg}{3kg+1kg}$

$\mu=\frac{3}{4}$

Given k =300 N/m.

Substitute the given values, we get

$\nu=\frac{1}{2\times3.14} \sqrt{\frac{300N/m}{3/4} }$

$\nu=\frac{1}{2\times3.14}\times20$

$\nu=3.18Hz$

   = 3 Hz.

Thus, the natural vibration frequency is  3 Hz.

#Learn More: Frequency in spring.

https://brainly.in/question/7838654

​