Question

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(ख) कहानी की उचित परिभाषा देते हुए उद्देश्य सम्बन्धी तत्त्व को स्पष्ट
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TOTT​

Answer 1

Answer:

Topic :- Oscillation

$\maltese\:\underline{\textsf{\textbf{AnsWer :}}}\:\maltese$

We are provided in the question that, a body of mass (m) is 2 kg attached to a spring is Displacement (x) through 0.04 m i.e Amplitude (A) = 0.04 m from its equilibrium position. The spring constant (k) is 200 N/m.

We need to find the following quantities :

Time period (T) = ?

Frequency (n) = ?

maximum Velocity = ?

maximum Acceleration = ?

First let's find the frequency (n) of the given body.

$\longrightarrow\:\:\sf n = \dfrac{1}{2\pi} \sqrt{ \dfrac{k}{m}}$

$\longrightarrow\:\:\sf n = \dfrac{1}{2\pi} \sqrt{ \dfrac{200}{2}}$

$\longrightarrow\:\:\sf n = \dfrac{1}{2\pi} \sqrt{ 100}$

$\longrightarrow\:\:\sf n = \dfrac{1}{2\pi} \sqrt{ 10 \times 10}$

$\longrightarrow\:\:\sf n = \dfrac{1}{2\pi} \times 10$

$\longrightarrow\:\:\sf n = \dfrac{10}{2\pi}$

$\longrightarrow\:\:\sf n = \dfrac{5}{\pi}$

$\longrightarrow\:\:\sf n = \dfrac{5}{ \dfrac{22}{7} }$

$\longrightarrow\:\:\sf n = \dfrac{5 \times 7}{ 22 }$

$\longrightarrow\:\:\sf n = \dfrac{35}{ 22 }$

$\longrightarrow\:\: \underline{ \boxed{\sf n \approx 1.59 \: Hz }}$

Now, we know that frequency is nothing but reciprocal of time period :

$\dashrightarrow\:\:\sf n = \dfrac{1}{T}$

$\dashrightarrow\:\:\sf 1.59= \dfrac{1}{T}$

$\dashrightarrow\:\:\sf T= \dfrac{1}{1.59}$

$\dashrightarrow\:\: \underline{ \boxed{\sf T=0.628 \: s}}$

For Calculating the maximum Velocity and maximum Acceleration first we need to find the angular velocity :

$:\implies \sf \omega = \dfrac{2\pi}{T}$

$:\implies \sf \omega = \dfrac{2\pi}{0.628}$

$:\implies \sf \omega = \dfrac{2 \times 3.14}{0.628}$

$:\implies \sf \omega = \dfrac{6.28}{0.628}$

$:\implies \underline{ \boxed{\sf \omega = 10 \: rad/s}}$

Now, let's calculate the maximum Velocity of the given particle.

$\longmapsto\:\:\sf V_{max} = A \omega$

$\longmapsto\:\:\sf V_{max} = 0.04 \times 10$

$\longmapsto\:\: \underline{ \boxed{\sf V_{max} = 0.4 \:m/s}}$

Now, we need to find the maximum Acceleration of the given particle.

$\leadsto\:\:\sf A_{max} = - A \omega^2$

$\leadsto\:\:\sf A_{max} = - 0.04 \times (10)^2$

$\leadsto\:\:\sf A_{max} = - 0.04 \times 100$

$\leadsto\:\:\sf A_{max} = - 0.04 \times 100$

$\leadsto\:\: \underline{ \boxed{\sf A_{max} = - 4 \: m/s^2}}$