If amplitude of oscillations of a simple pendulum is increased by 20%,
then percentage change in its maximum kinetic energy will be
20%
70%
40%
44%
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If amplitude of oscillations of a simple pendulum is increased by 20%,
then percentage change in its maximum kinetic energy will be
20%
70%
40%
44%
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Answer:
☆Concept:
》kinetic energy of the SHM is highest at its mean position.
where velocity can written as ...
[tex]\ V_{max} = \omega A[/tex], and [tex]\ KE= \dfrac{1}{2}mv^{2}[/tex]
so,[tex]\boxed{KE = \dfrac{1}{2}m \omega^{2}A^{2}}[/tex]
--------
☆Solution:
• [tex]\ KE = \dfrac{1}{2}m \omega^{2}A^{2}[/tex]
• here let's take, [tex] \dfrac{1}{2}m \omega^{2} = constant =k[/tex]
• [tex]\ KE = kA^{2}[/tex]
-----
》Inital amplitude([tex]\ A_{i}[/tex]) = A
》final amplitude([tex]\ A_{f}[/tex]) = [tex]\ A+ \dfrac{20}{100}A[/tex]
》final amplitude([tex]\ A_{f}[/tex]) = 6A/5
------
》[tex]\ KE_{i} = kA^{2}[/tex]
》[tex]\ KE_{f} = k( \dfrac{6A}{5})^{2}[/tex]
》[tex]\ KE_{f} = k \dfrac{36A^{2}}{25}[/tex]
------
% change in KE = [tex]\dfrac{ K_{f} - K_{i}}{K_{i}} × 100[/tex]
》 change = ([tex] \dfrac{ k \dfrac{36A^{2}}{25} - kA^{2}}{kA^{2}} × 100[/tex])%
》change = ([tex] \dfrac{ \dfrac{36}{25} - 1}{1} × 100[/tex])%
》change =([tex] \dfrac{36-25}{25}× 100[/tex])%
》change = (11 × 4)% = 44%
so change in kinetic energy is 44%
Given info : If amplitude of oscillations of a simple pendulum is increased by 20%.
To find : The percentage change in its maximum kinetic energy will be...
solution : kinetic energy of oscillation is directly proportional to square of its amplitude.
i.e., K.E_i = kA² ...(1)
where k is proportionality constant and A is amplitude.
now A is increased by 20%
new amplitude, A' = A + 20% of A = 1.2A
now K.E_f = k(1.2A)² = 1.44kA² ...(2)
now percentage change in kinetic energy = (K.E_f - K.E_i)/K.E_i × 100
= (1.44kA² - kA²)/kA² × 100
= 44 %
Therefore the percentage change in its kinetic energy is 44 %.
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