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Home » A body describes simple harmonic motion with an amplitude of and a period of . Find the acceleration and velocity of the body when the displacement is (a) (b)
A body describes simple harmonic motion with an amplitude of 5 \mathrm{~cm} and a period of 0.2 \mathrm{~s}. Find the acceleration and velocity of the body when the displacement is
(a) 5 \mathrm{~cm}
(b) 3 \mathrm{~cm}
CBSE | Class 11 | NCERT | Oscillations | Physics
A body describes simple harmonic motion with an amplitude of 5 \mathrm{~cm} and a period of 0.2 \mathrm{~s}. Find the acceleration and velocity of the body when the displacement is
(a) 5 \mathrm{~cm}
(b) 3 \mathrm{~cm}

Amplitude is given as =5 \mathrm{~cm}=0.05 \mathrm{~m}

Time period is given as =0.2 \mathrm{~s}

When the displacement is y, then

acceleration is given as A=-\omega^{2} y

Velocity is given as v=\omega \sqrt{r^{2}-y^{2}}

=2 \pi / 0.2=10 \pi \mathrm{rad} / \mathrm{s}

(a) When the displacement is y=5 \mathrm{~cm}=0.05 \mathrm{~m},

Acceleration will be A=-(10 \pi)^{2}(0.05)=5 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}

(b) When the displacement is y=3 \mathrm{~cm}=0.03 \mathrm{~m},

Acceleration will be A=-(10 \pi)^{2}(0.03)=3 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}

Velocity will be V=10 \pi \sqrt{(0.05)^{2}-(0.03)^{2}}=10 \pi \times 0.04=0.4 \pi \mathrm{m} / \mathrm{s}

i

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