Question

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%​

Answer 1

Answer:

☆Concept:

》kinetic energy of the SHM is highest at its mean position.

where velocity can written as ...

$\ V_{max} = \omega A$, and $\ KE= \dfrac{1}{2}mv^{2}$

so,$\boxed{KE = \dfrac{1}{2}m \omega^{2}A^{2}}$

--------

☆Solution:

• $\ KE = \dfrac{1}{2}m \omega^{2}A^{2}$

• here let's take, $\dfrac{1}{2}m \omega^{2} = constant =k$

• $\ KE = kA^{2}$

-----

》Inital amplitude($\ A_{i}$) = A

》final amplitude($\ A_{f}$) = $\ A+ \dfrac{20}{100}A$

》final amplitude($\ A_{f}$) = 6A/5

------

》$\ KE_{i} = kA^{2}$

》$\ KE_{f} = k( \dfrac{6A}{5})^{2}$

》$\ KE_{f} = k \dfrac{36A^{2}}{25}$

------

% change in KE = $\dfrac{ K_{f} - K_{i}}{K_{i}} × 100$

》 change = ($\dfrac{ k \dfrac{36A^{2}}{25} - kA^{2}}{kA^{2}} × 100$)%

》change = ($\dfrac{ \dfrac{36}{25} - 1}{1} × 100$)%

》change =($\dfrac{36-25}{25}× 100$)%

》change = (11 × 4)% = 44%

so change in kinetic energy is 44%

Answer 2

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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