Answer
Verified
38.1k+ views
Hint: The spring constant of the spring is inversely proportional to the length of the spring. If the same spring is divided into smaller spring lengths then the resulting spring will have more stiffness than the original one. We find the equivalent stiffness of the combination then use it in the formula of the period of spring-mass system.
Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Where T is the period, k is the spring constant and m is the mass of the block.
Complete step by step solution:
Figure: The spring-mass system
For the initial case, using period formula,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
As we know that the spring constant of the spring is inversely proportional to the length of the spring, so when the spring is divided into two equal parts, then the spring constant of each equal parts will be twice of the original spring.
It is given that the spring constant of original spring is k, so the spring constant of each part will be 2k. Then, the equivalent spring constant in parallel is,
\[{K_{eq}} = 2k + 2k = 4k\]
So, the final period of the system will be,
\[{T_2} = 2\pi \sqrt {\dfrac{m}{{4K}}} \]
\[{T_2} = \dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)\]
On dividing the expression for the periods, we get
\[\dfrac{T}{{{T_2}}} = \dfrac{{\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}{{\dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}\]
\[\dfrac{T}{{{T_2}}} = 2\]
\[{T_2} = \dfrac{T}{2}\]
So, the period of the new spring-mass system is half of the initial period.
Therefore, the correct option is (D).
Note: We should assume that the spring’s mass is negligible in relative to the mass suspended through the spring. Otherwise the motion of the block will not be perfect simple harmonic.
Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Where T is the period, k is the spring constant and m is the mass of the block.
Complete step by step solution:
Figure: The spring-mass system
For the initial case, using period formula,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
As we know that the spring constant of the spring is inversely proportional to the length of the spring, so when the spring is divided into two equal parts, then the spring constant of each equal parts will be twice of the original spring.
It is given that the spring constant of original spring is k, so the spring constant of each part will be 2k. Then, the equivalent spring constant in parallel is,
\[{K_{eq}} = 2k + 2k = 4k\]
So, the final period of the system will be,
\[{T_2} = 2\pi \sqrt {\dfrac{m}{{4K}}} \]
\[{T_2} = \dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)\]
On dividing the expression for the periods, we get
\[\dfrac{T}{{{T_2}}} = \dfrac{{\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}{{\dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}\]
\[\dfrac{T}{{{T_2}}} = 2\]
\[{T_2} = \dfrac{T}{2}\]
So, the period of the new spring-mass system is half of the initial period.
Therefore, the correct option is (D).
Note: We should assume that the spring’s mass is negligible in relative to the mass suspended through the spring. Otherwise the motion of the block will not be perfect simple harmonic.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Other Pages
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
A point charge q placed at the point A is A In stable class 12 physics JEE_Main