Question

A particle executes SHM given by x = 0.24 cos (400t - 0.5) in SI units. Find the amplitude, frequency, and period of vibration.

Answer 1

AnswEr :

• Amplitude = 0.24 m

• Time Period = 1.57 × 10^-2 s

• Frequency = 63.69 Hz

The SHM of the particle is defined as :

$\sf x = 0.24cos(400t - 0.5)----------(1)$

General Equation of SHM :

$\sf x = Acos(\omega T + \phi)----------(2)$

Comparing equations (1) and (2),

• Amplitude = 0.24 m

• Angular Frequency = 400 rad/s

• Time Period = t s

Now,

Time period is given as :

$\sf \: t = \dfrac{2\pi}{ \omega} \\ \\ \dashrightarrow \: \sf \: t = \dfrac{ 2\times 3.14}{400} \\ \\ \dashrightarrow \: \sf \: t = \dfrac{2 \times 3.14 \times {10}^{ - 2} }{4} \\ \\ \dashrightarrow \: \boxed{ \boxed{ \sf \: t = 1.57 \times {10}^{ - 2} \: s}}$

Frequency is given as :

$\sf \: f = \dfrac{1}{t} \\ \\ \longrightarrow \: \sf \: f = \dfrac{1}{1.57 \times {10}^{ - 2} } \\ \\ \longrightarrow \: \sf \: f = \dfrac{100}{1.57} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf{f = 63.69 \: {s}^{ - 1} }}}$

Answer 2

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Question :

A particle executes SHM given by x = 0.24 cos (400t - 0.5) in SI units. Find the amplitude, frequency, and period of vibration.

Solution :

$</p><p>\tt {\orange {Step-By-Step-Explaination \: : }}$

$\tt {\red{ Given \: - }}$

• Amplitude = 0.24 m

• Angular Frequency = 400 rad/s

• Time Period = t s

Substituting The Above Values -

$\begin{lgathered}\sf \: T.P = \dfrac{2\pi}{ \omega} \\ \\ \dashrightarrow \: \sf \: T.P = \dfrac{ 2\times 3.14}{400} \\ \\ \dashrightarrow \: \sf \: T.P = \dfrac{2 \times 3.14 \times {10}^{ - 2} }{4} \\ \\ \dashrightarrow \: \boxed{ \boxed{ \implies \: \sf \:Time \: period = 1.57 \times {10}^{ - 2} \: s}}\end{lgathered}$

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$\begin{lgathered}\sf \: f = \dfrac{1}{t} \\ \\ \longrightarrow \: \sf \: f = \dfrac{1}{1.57 \times {10}^{ - 2} } \\ \\ \longrightarrow \: \sf \: f = \dfrac{100}{1.57} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf{f = 63.69 \: {s}^{ - 1} }}}\end{lgathered}$

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