
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
233.4k+ 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.
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