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A block of mass 0.1kg is connected to an elastic spring of spring constant \[640N{m^{ - 1}}\] and oscillates in a damping medium of damping constant \[{10^{ - 2}}k{g_{}}{s^{ - 1}}\]. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to:
(A) 5s
(B) 7s
(C) 3.5s
(D) 2s

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Last updated date: 13th Jun 2024
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Answer
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Hint For finding the time taken for energy to drop half its initial value, first consider the overall amplitude of damped oscillation given as\[A = {A_0}.{e^{\dfrac{{ - bt}}{{2m}}}}\]. Find the relation when energy drops half the value and substitute again in this equation to obtain its original t value.

Complete Step By Step Solution
Amplitude of an damped oscillation is given using the equation, \[A = {A_0}.{e^{\dfrac{{ - bt}}{{2m}}}}\]. It is known that energy of the damped oscillation system is directly related to its amplitude. It is said that since the system dissipates energy gradually with respect to time, the amplitude will decrease with time.
Now, work done is defined as a product of force and displacement. Here, the force acting on the mass is due to spring. Hence , \[W = F \times {x^2}\], which means that work done is directly proportional to square of displacement. In graphical terms, displacement in waveforms is measured as amplitude. Thus energy is proportional to square of amplitude.
When energy is halved , \[\dfrac{E}{2} \propto {a_{new}}^2\]. This implies
\[ \Rightarrow \sqrt {\dfrac{E}{2}} \propto {a_{new}}\]
\[ \Rightarrow \dfrac{{{a_0}}}{{\sqrt 2 }} \propto {a_{new}}\]
Substituting the value in the amplitude equation we get,
\[{A_{new}} = {A_0}.{e^{\dfrac{{ - bt}}{{2m}}}}\]
\[ \Rightarrow \dfrac{{{A_{_0}}}}{{\sqrt 2 }} = {A_0}.{e^{\dfrac{{ - bt}}{{2m}}}}\](since we know that \[{A_{_{new}}} = {A_0}/\sqrt 2 \]which was derived above)
Cancelling the like term we get
\[ \Rightarrow \dfrac{1}{{\sqrt 2 }} = {e^{\dfrac{{ - bt}}{{2m}}}}\]
We know that damping constant b is \[{10^{ - 2}}\] and mass m as 0.1 kgs. Substituting this we get,
\[ \Rightarrow \dfrac{1}{{\sqrt 2 }} = {e^{\dfrac{{ - {{10}^{ - 2}}t}}{{2 \times 0.1}}}}\]
Removing \[e\] by multiplying \[\ln \]on both sides we get,
\[ \Rightarrow \ln (\dfrac{1}{{\sqrt 2 }}) = \dfrac{{ - {{10}^{ - 2}}t}}{{0.2}}\]
\[ \Rightarrow - 0.35 = \dfrac{{ - {{10}^{ - 2}}t}}{{0.2}}\]
\[ \Rightarrow 0.35 \times 20 = t\]
\[t = 7s\]

Hence, Option (d) is the right answer for the given question.

Note Amplitude of a wave is defined as the distance from the central line to the top of an upward crest or to the bottom of a downward trough. Crest is the maximum point of the wave and trough is the minimum point of the wave.