A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1kg, the time period of oscillations becomes 5s. The value of m in kg is
(A) $\dfrac{9}{{16}}$
(B) $\dfrac{3}{4}$
(C) $\dfrac{4}{3}$
(D) $\dfrac{{16}}{9}$
Answer
Verified
116.7k+ views
Hint: This is a question of simple harmonic motion in which the restoring force acting on the body is proportional directly to the displacement of the body. So, we can use the time period expression of simple harmonic motion and calculate the mass.
Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Complete step by step answer
In the question, it says that the mass attached to the lower end of spring is pulled towards right, then spring gets stretched and a restoring force acts in the opposite direction (left) due to elasticity. When the mass is stretched towards the left, the spring compresses in the left direction and so the restoring force acts towards the right direction. When we release mass, it will oscillate to and fro because of the restoring forces about its equilibrium position. So, it performs a simple harmonic motion. In simple harmonic motion, the acceleration is proportional to displacement and acts in the opposite direction from it.
We know that the time period T in a simple harmonic motion is given by
\[T = 2\pi \sqrt {\dfrac{m}{k}} \] here we can see that T is proportional to root of m
Or, we can say $\dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
Here ${T_1} = 3\sec $${T_2} = 5\sec $${m_1} = m$ and ${m_2} = m + 1$
Putting these values, we get
$\dfrac{3}{5} = \sqrt {\dfrac{m}{{m + 1}}} \Rightarrow m = \dfrac{9}{{16}}kg$this is the value of m
Hence, the correct answer is A.
Note:
In simple harmonic motion, a particle has both potential and kinetic energy. It can be proved in the following manner. Say, when a particle is displaced from equilibrium position, it possesses potential energy and because of presence of velocity it has kinetic energy. At maximum displacement, the total energy is in the form of potential energy that is kinetic energy is zero and at the equilibrium position, total energy is in kinetic energy form which means potential energy is zero.
Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Complete step by step answer
In the question, it says that the mass attached to the lower end of spring is pulled towards right, then spring gets stretched and a restoring force acts in the opposite direction (left) due to elasticity. When the mass is stretched towards the left, the spring compresses in the left direction and so the restoring force acts towards the right direction. When we release mass, it will oscillate to and fro because of the restoring forces about its equilibrium position. So, it performs a simple harmonic motion. In simple harmonic motion, the acceleration is proportional to displacement and acts in the opposite direction from it.
We know that the time period T in a simple harmonic motion is given by
\[T = 2\pi \sqrt {\dfrac{m}{k}} \] here we can see that T is proportional to root of m
Or, we can say $\dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{m_1}}}{{{m_2}}}} $
Here ${T_1} = 3\sec $${T_2} = 5\sec $${m_1} = m$ and ${m_2} = m + 1$
Putting these values, we get
$\dfrac{3}{5} = \sqrt {\dfrac{m}{{m + 1}}} \Rightarrow m = \dfrac{9}{{16}}kg$this is the value of m
Hence, the correct answer is A.
Note:
In simple harmonic motion, a particle has both potential and kinetic energy. It can be proved in the following manner. Say, when a particle is displaced from equilibrium position, it possesses potential energy and because of presence of velocity it has kinetic energy. At maximum displacement, the total energy is in the form of potential energy that is kinetic energy is zero and at the equilibrium position, total energy is in kinetic energy form which means potential energy is zero.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids