# If a spring has a period $T$, and is cut into the $n$ equal parts, then the period of each part will be:

A. $T\sqrt n $

B. $\dfrac{T}{{\sqrt n }}$

C. $nT$

D. $T$

Answer

Verified

236.1k+ views

**Hint:**The spring constant is always in the linear range. At the lower gravity, the period did not change, The system’s equilibrium position gets different. Hence the period fully depends on the spring constant and mass of the object.

**Complete step by step solution:**

The hooke's law execute a simple harmonic motion as follows

The period $T$ , mass $m$ , and spring of spring constant \[K\] is given by the equation as follows,

$T = 2\pi \sqrt {\dfrac{m}{k}} $ ………(1)

After the spring is cut into four equal parts, let \[K\] be the spring constant of little spring. Hence all parts are equal we resume all small springs following Hooke's Law and have equal spring constant.

Where,

\[K\] is known as the proportionality constant and which is also called a spring constant. The spring constant \[K\] depends on the factors that are listed below,

Nature of the material of the spring,

Cross-Sectional Area of the spring,

Length of the spring.

For a spring,

$T = 2\pi \sqrt {\dfrac{m}{k}} $

For every piece of spring, the constant is $nk$

$T' = 2\pi \sqrt {\dfrac{m}{{nk}}} $

Rearrange the above equation and simplify the equation we get,

$T' = 2\pi \sqrt {\dfrac{m}{k} \times } \dfrac{1}{{\sqrt n }}$

Hence the period of each part is, $T' = \dfrac{T}{{\sqrt n }}$

**So, the correct answer is, Option (B)**

**Note:**

The spring constant and mass help to determine the period.

The equilibrium position of the system is determined by the gravitational force.

The equilibrium position does not depend on gravity.

Last updated date: 25th Sep 2023

•

Total views: 236.1k

•

Views today: 3.36k