Question

A mass on a spring oscillates along a vertical line, taking 12s to complete 10 oscillations.

Calculate the (a)time period, and (b) the angular frequency.
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Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• A mass on a spring oscillates along a vertical line taking 12s to complete 10 oscillation.

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Time Period (T)
• Angular Frequency ${\sf (\omega)}$

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$\color{darkblue}\underline{\underline{\sf Formula \: Used-}}$

$\color{violet}\bullet\underline{\boxed{\sf Time\: Period (T)=\dfrac{Total\:Time}{No.\:of\: oscillating}}}$

$\color{violet}\bullet\underline{\boxed{\sf Angular\: Frequency (\omega)= \dfrac{2π}{T}}}$

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❏ Time Period

$\implies{\sf T =\dfrac{12}{10}}$

$\color{red}\implies{\sf T=1.2\:s}$

❏ Angular Frequency

$\large\implies{\sf \omega=\dfrac{2π}{T}}$

$\implies{\sf \omega =\dfrac{2×3.12}{1.2} }$

$\implies{\sf \omega =\dfrac{6.28}{1.2} }$

$\color{red}\implies{\sf Angular \: Frequency (\omega)=5.23\:rad/s }$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Angular Frequency $\color{red}{\sf 5.23\:rad/s}$

Time period $\color{red}{\sf 1.2\:sec}$

Answer 2

Solution :

⏭ Given:

✏ No. of oscillations = 10

✏ Time interval = 12s

⏭ To Find:

• Time period
• Angular frequency

⏭ Concept:

✏ Time period is defined as time requires to complete one oscillation.

⏭ Formula:

• Time period

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{Time \: period = \frac{Total \: time}{No. \: of \: oscillations}}}}}} \: \star$

• Angular frequency

$\star \: \underline{ \boxed{ \bold{ \sf{ \large{ \purple{ \omega = \frac{2\pi}{T}}}}}}} \: \star$

⏭ Calculation:

• Time period

$\implies \sf \: T = \dfrac{12}{10} \\ \\ \implies \: \boxed{ \bold{ \red{\sf{T = 1.2 \: s}}}}$

• Angular frequency

$\implies \sf \: \omega = \dfrac{2\pi }{1.2} \\ \\ \implies \sf \: \omega = \frac{6.28}{1.2} \\ \\ \implies \: \boxed{ \bold{ \sf{ \green{ \omega = 5.23 \: rad {(s)}^{ - 1}}}}}$