
The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be?
A) $\dfrac{{\text{T}}}{4}$
B) ${\text{T}}$
C) $\dfrac{{\text{T}}}{2}$
D) $2{\text{T}}$
Answer
232.8k+ views
Hint: For solving this question we must be familiar with Hooke’s law and the mathematical representation which is given by ${\text{T}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{k}}$, where T is the time period, m is the mass and k is spring constant.
Complete step by step answer:
We know that motion of a spring for a given mass is described as Simple Harmonic Motion (SHM). We should also know that spring constant is the measure of how stiff a spring really is so greater the value of spring constant more force is needed to pull the spring. Equation of motion for one-dimensional simple harmonic motion can be obtained by newton’s second laws and Hooke’s law which is
${{\text{F}}_{net}}\, = \,m\dfrac{{{d^2}x}}{{d{t^2}}}\, = \, - kx$ where m is the mass of the body, x is the displacement and k is the spring constant.
Given from the question, T is the time period of spring with mass suspended as m, so from given formula of hint we can write it as
${\text{T}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{k}} $ --(equation 1)
Now we know that when spring is cut into ‘n’ equal parts, new spring constant(k`) becomes
= $k({\text{original spring constant)}}\, \times \,{\text{equal number of parts(n)}}\,{\text{ = }}\,n\, \times \,k$
In this case since spring is cut into 4 equal parts so new spring constant (k`) = $4k$
Let’s assume new time period as T`, so we can write T` as
${\text{T`}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{{k`}}} $
Substituting value of k` in above equation we can write
${\text{T`}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{{4k}}} $ --(equation 2)
Dividing equation 2 by 1 we get
$\dfrac{{{\text{T`}}}}{{\text{T}}}\, = \,\dfrac{{{\text{2}}\pi \sqrt {\dfrac{m}{{4k}}} }}{{{\text{2}}\pi \sqrt {\dfrac{m}{k}} }}$
$\dfrac{{{\text{T`}}}}{{\text{T}}}\, = \,\dfrac{1}{2}$
So finally, new time period (T`) can be written as
${\text{T`}}\,{\text{ = }}\,\dfrac{{\text{T}}}{2}$
Hence, we can say that the new time period will be half of the original time period.
Therefore, the correct option is (C).
Note: Hooke’s law is only valid till proportional limit of the material, once the body is deformed permanently Hooke’s laws of elasticity can’t be applied. Students sometimes get confused between elastic limit and proportional limit but these two are different, proportional limit occurs ahead of elastic limit.
Complete step by step answer:
We know that motion of a spring for a given mass is described as Simple Harmonic Motion (SHM). We should also know that spring constant is the measure of how stiff a spring really is so greater the value of spring constant more force is needed to pull the spring. Equation of motion for one-dimensional simple harmonic motion can be obtained by newton’s second laws and Hooke’s law which is
${{\text{F}}_{net}}\, = \,m\dfrac{{{d^2}x}}{{d{t^2}}}\, = \, - kx$ where m is the mass of the body, x is the displacement and k is the spring constant.
Given from the question, T is the time period of spring with mass suspended as m, so from given formula of hint we can write it as
${\text{T}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{k}} $ --(equation 1)
Now we know that when spring is cut into ‘n’ equal parts, new spring constant(k`) becomes
= $k({\text{original spring constant)}}\, \times \,{\text{equal number of parts(n)}}\,{\text{ = }}\,n\, \times \,k$
In this case since spring is cut into 4 equal parts so new spring constant (k`) = $4k$
Let’s assume new time period as T`, so we can write T` as
${\text{T`}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{{k`}}} $
Substituting value of k` in above equation we can write
${\text{T`}}\,{\text{ = }}\,{\text{2}}\pi \sqrt {\dfrac{m}{{4k}}} $ --(equation 2)
Dividing equation 2 by 1 we get
$\dfrac{{{\text{T`}}}}{{\text{T}}}\, = \,\dfrac{{{\text{2}}\pi \sqrt {\dfrac{m}{{4k}}} }}{{{\text{2}}\pi \sqrt {\dfrac{m}{k}} }}$
$\dfrac{{{\text{T`}}}}{{\text{T}}}\, = \,\dfrac{1}{2}$
So finally, new time period (T`) can be written as
${\text{T`}}\,{\text{ = }}\,\dfrac{{\text{T}}}{2}$
Hence, we can say that the new time period will be half of the original time period.
Therefore, the correct option is (C).
Note: Hooke’s law is only valid till proportional limit of the material, once the body is deformed permanently Hooke’s laws of elasticity can’t be applied. Students sometimes get confused between elastic limit and proportional limit but these two are different, proportional limit occurs ahead of elastic limit.
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