Question

The time period of a body suspended from a spring is $T$. What will be the new time period, if the spring is cut into two equal parts and (ii) the same mass is suspended from one part, (ii) the mass is suspended simultaneously from both the parts in parallel ?
A. $\dfrac{T}{{2\sqrt 2 }},\dfrac{T}{2}$
B. $\dfrac{T}{{\sqrt 2 }},T$
C. $\sqrt 2 T,\dfrac{T}{2}$
D. $\dfrac{T}{{\sqrt 2 }},\dfrac{T}{2}$

Answer 1

Hint: In order to solve this question we need to understand hooke's law.According to hooke's law the force on a spring is constant and it is directly proportional to the displacement and acts opposite to it. Removing the proportionality, the magnitude of a force on mass attached with spring is equal to the product of stiffness or spring constant and displacement of body.

Complete step by step answer:
Consider a spring of spring constant “k” and let us assume that mass “m” is attached to its one end and also let the length of spring be $l$. Then the mass oscillates and its motion is known as simple harmonic motion. Let the time period of SHM be “T”. Then from formula of SHM we know,
$T = 2\pi \sqrt {\dfrac{m}{k}} $
So, Magnitude of Force on spring when it is not cut, is
$\left| {\vec F} \right| = kl \to (i)$
So when the spring is cut into two equal parts, then the magnitude of force on one spring would be,
$\left| {\vec F'} \right| = k'\dfrac{l}{2} \to (ii)$
We know, force on spring on spring always remains constant, equating equation (i) and (ii) we get,
$kl = k'\dfrac{l}{2}$
$\Rightarrow k' = 2k$

(a) When the mass is suspended from one part, the new time period would be, $T' = 2\pi \sqrt {\dfrac{m}{{k'}}} $
Putting values we get,
$T' = \dfrac{1}{{\sqrt 2 }}(2\pi \sqrt {\dfrac{m}{k}} )$
Since we know,
$T = 2\pi \sqrt {\dfrac{m}{k}} $
$\therefore T' = \dfrac{T}{{\sqrt 2 }}$

(b) When the two parts are connected in parallel, then net effective stiffness constant would be,
$k'' = k' + k'$
Putting values we get,
$k'' = 2k'$
$\Rightarrow k'' = 4k$
So the new time period would be,
$T'' = 2\pi \sqrt {\dfrac{m}{{k''}}} $
Putting values we get,
$T'' = 2\pi \sqrt {\dfrac{m}{{4k}}} $
$\Rightarrow T'' = \dfrac{1}{2}(2\pi \sqrt {\dfrac{m}{k}} )$
Since we know,
$T = 2\pi \sqrt {\dfrac{m}{k}} $
$\therefore T'' = \dfrac{T}{2}$

So the correct option is D.

Note: It should be remembered that although the suspending force of gravity acts on mass, this would not change the time period of SHM as it is only a constant force. SHM motion is defined as Simple Harmonic Motion, in which a body oscillates with a time period and the force causing oscillation acts in the opposite direction of displacement.