
A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $\dfrac{\text{2A}}{\text{3}}$ from equilibrium position. The new amplitude of the motion is:
(A). $\dfrac{\text{A}}{\text{3}}\sqrt{\text{41}}$
(B). 3A
(C). $\text{A}\sqrt{\text{3}}$
(D). $\dfrac{\text{7A}}{\text{3}}$
Answer
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Hint: In order to solve this question, we have to have an idea about the simple harmonic motion. Once we have an idea we will be able to solve the question by applying the formula of amplitude.
Complete step-by-step answer:
Some specific properties of Simple Harmonic Motion or SHM include:
1. The motion of the body performing the Simple Harmonic Motion, is periodic which means the action repeats itself at constant frequency.
2. The velocity of the body is maximum when the body passes the equilibrium. And the velocity of the body is zero when the body passes through extreme positions in the motion.
We know that the formula for velocity of Simple Harmonic Motion is written as,
$\text{V = }\sqrt{{{\text{A}}^{\text{2}}}\text{ }-\text{ }{{\text{x}}^{\text{2}}}}$
Here, in the question it is given that,
The amplitude of the particle undergoing SHM = A
The distance of the particle from the equilibrium position = $\dfrac{\text{2A}}{3}$
Therefore, the velocity of that body can be written as,
$\text{V = }\sqrt{{{\text{A}}^{\text{2}}}\text{ }-\text{ }{{\text{x}}^{\text{2}}}}$
$\Rightarrow \text{ V = }\sqrt{{{\text{A}}^{\text{2}}}\text{ }-\text{ }\dfrac{\text{4}{{\text{A}}^{\text{2}}}}{9}}$
$\Rightarrow \text{ V = }\sqrt{\dfrac{\text{5}{{\text{A}}^{\text{2}}}}{\text{9}}}\text{ }\ldots \text{(i)}$
Also,
Given in the question, when the amplitude of motion is changed to $\text{{A}'}$, the speed of SHM triples.
Hence, we can write the equation as,
$\text{3V = }\sqrt{{{{\text{{A}'}}}^{\text{2}}}\text{ }-\text{ }\dfrac{4{{\text{A}}^{\text{2}}}}{9}}\text{ }\ldots \text{(ii)}$
Dividing equations (i) and (ii),
We get,
$\text{{A}' = }\dfrac{\text{7A}}{3}$
Therefore, the correct option is Option D.
Note: Simple Harmonic Motion is considered as a periodic motion, where the restoring force on the moving object is directly proportional to the magnitude of displacement of the object along with the act towards the equilibrium position of the object. In case of Simple Harmonic Motion, the acceleration is towards a fixed point in a line, which is directly proportional to the distance of the body from that particular point.
Complete step-by-step answer:
Some specific properties of Simple Harmonic Motion or SHM include:
1. The motion of the body performing the Simple Harmonic Motion, is periodic which means the action repeats itself at constant frequency.
2. The velocity of the body is maximum when the body passes the equilibrium. And the velocity of the body is zero when the body passes through extreme positions in the motion.
We know that the formula for velocity of Simple Harmonic Motion is written as,
$\text{V = }\sqrt{{{\text{A}}^{\text{2}}}\text{ }-\text{ }{{\text{x}}^{\text{2}}}}$
Here, in the question it is given that,
The amplitude of the particle undergoing SHM = A
The distance of the particle from the equilibrium position = $\dfrac{\text{2A}}{3}$
Therefore, the velocity of that body can be written as,
$\text{V = }\sqrt{{{\text{A}}^{\text{2}}}\text{ }-\text{ }{{\text{x}}^{\text{2}}}}$
$\Rightarrow \text{ V = }\sqrt{{{\text{A}}^{\text{2}}}\text{ }-\text{ }\dfrac{\text{4}{{\text{A}}^{\text{2}}}}{9}}$
$\Rightarrow \text{ V = }\sqrt{\dfrac{\text{5}{{\text{A}}^{\text{2}}}}{\text{9}}}\text{ }\ldots \text{(i)}$
Also,
Given in the question, when the amplitude of motion is changed to $\text{{A}'}$, the speed of SHM triples.
Hence, we can write the equation as,
$\text{3V = }\sqrt{{{{\text{{A}'}}}^{\text{2}}}\text{ }-\text{ }\dfrac{4{{\text{A}}^{\text{2}}}}{9}}\text{ }\ldots \text{(ii)}$
Dividing equations (i) and (ii),
We get,
$\text{{A}' = }\dfrac{\text{7A}}{3}$
Therefore, the correct option is Option D.
Note: Simple Harmonic Motion is considered as a periodic motion, where the restoring force on the moving object is directly proportional to the magnitude of displacement of the object along with the act towards the equilibrium position of the object. In case of Simple Harmonic Motion, the acceleration is towards a fixed point in a line, which is directly proportional to the distance of the body from that particular point.
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