Question

The maximum velocity of a particle executing simple
harmonic motion is 100 cm s- and the maximum
acceleration is 157 cm s. Determine the periodic-
time.
Ans. 4 s.​

Answer 1

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Answer 2

Answer:

• Time Period (T) is 4 Seconds.

Given:

1. Maximum velocity = 100 cm/s
2. Maximum Acceleration = 157 cm/s.

Explanation:

$\rule{300}{1.5}$

From Maximum velocity Formula

$\large\bigstar \: {\boxed{\tt v_{max} = \omega A}}$

$\bold{Here}\begin{cases}\text{A Denotes Amplitude} \\ \omega \text{ Denotes Angular Frequency}\end{cases}$

Now,

$\large{\boxed{\tt v_{max} = \omega A}}$

Substituting the values,

$\longmapsto \large{\tt 100 = \omega A}$

$\longmapsto \large{\tt 100 = \omega A \: ----(1)}$

From Maximum Acceleration Formula.

$\large\bigstar \: {\boxed{\tt a_{max} = \omega^2 A}}$

$\bold{Here}\begin{cases}\text{A Denotes Amplitude} \\ \omega \text{ Denotes Angular Frequency}\end{cases}$

Now,

$\large{\boxed{\tt a_{max} = \omega^2 A}}$

Substituting the values,

$\longmapsto \large{\tt 157 = \omega^2 A}$

$\longmapsto \large{\tt 157 = \omega^2 A \: ----(2)}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Dividing Equation (2) by (1) I.e. [Equation-2/Equation-1]

$\longmapsto \large{\tt \dfrac{157}{100} = \dfrac{\omega^2 A}{\omega A}}$

$\longmapsto \large{\tt \dfrac{157}{100} = \cancel{\dfrac{\omega^2 A}{\omega A}}}$

$\longmapsto \large{\tt \dfrac{157}{100} = \omega}$

$\longmapsto \large{\tt \omega = \dfrac{157}{100}}$

$\longmapsto\large{\underline{\boxed{\tt \omega = 1.57 \: rad/sec}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Formula we Know,

$\large\bigstar \: {\boxed{\tt \omega = \dfrac{2 \pi}{T}}}$

$\bold{Here}\begin{cases}\text{T Denotes Time Period} \\ \omega \text{ Denotes Angular Frequency}\end{cases}$

Now,

$\large{\boxed{\tt \omega = \dfrac{2 \pi}{T}}}$

Substituting the values,

$\longmapsto\large{\tt 1.57 = \dfrac{2 \times 3.14}{T}}$

$\longmapsto\large{\tt 1.57 \times T= 2 \times 3.14}$

$\longmapsto\large{\tt T = \dfrac{2 \times 3.14}{1.57}}$

$\longmapsto\large{\tt T = \cancel{\dfrac{2 \times 3.14}{1.57}}}$

$\longmapsto\large{\tt T = 2 \times 2}$

$\longmapsto\large{\underline{\boxed{\red{\tt T = 4 \: Second}}}}$

∴ Periodic Time Period (T) is 4 Seconds.

$\rule{300}{1.5}$