Question

The force-constant of an ideal spring is 200 Nm^-1.
A body of mass (200/pie^2) kg is suspended from it
and is made to oscillate. Find the time period of the
oscillation.
​

Answer 1

Answer :

• Hence, the time period of the oscillation is 2 s.

Explanation :

Given :

• Force constant of the spring, K = 200 N/m
• Mass of the body, m = 200/π² kg

To find :

• The time period of the oscillation, t = ?

Knowledge required :

Formula for time period in case of a spring :

⠀⠀⠀⠀⠀⠀⠀⠀⠀T = 2π√(m/k)⠀

Where :

• T = Time period.
• m = Mass of the body.
• k = Spring constant.

Solution :

By using the formula for Time Period in case of a string, we get :

⠀⠀=> T = 2π√(m/k)

⠀⠀=> T = 2π√(200/π²/200)

⠀⠀=> T = 2π√[200/(200 × π²)]

⠀⠀=> T = 2π√(1/π²)

⠀⠀=> T = 2π × 1/π

⠀⠀=> T = 2

⠀⠀⠀⠀⠀⠀⠀∴ T = 2 s

Therefore,

• The time period of the oscillation, t = 2 s.

Answer 2

Answer:

Given :-

• The force constant of an ideal spring is 200 Nm-¹. A body of mass is 200/π² kg is suspended from it and to oscillate.

To Find :-

• What is the time period of the oscillation.

Formula Used :-

${\red{\boxed{\large{\bold{T =\: 2{\pi}\sqrt{\dfrac{m}{k}}}}}}}$

where,

• T = Time Period
• m = Mass
• k = Spring Constant

Solution :-

Given :

• Mass = $\sf \dfrac{200}{{\pi}^{2}}$
• Spring Constant = 200 Nm-¹

According to the question by using the formula we get,

⇒ $\sf T =\: 2{\pi}\sqrt\dfrac{\cancel{200}}{{\pi}^{2} \times \cancel{200}}$

⇒ $\sf T =\: 2{\pi}\sqrt\dfrac{1}{{\pi}^{2}}$

⇒ $\sf T =\: \dfrac{2\cancel{{\pi}}}{\cancel{{\pi}}}$

⇒ $\sf T =\: \dfrac{2}{1}$

➠ $\sf\bold{\purple{T =\: 2\: seconds}}$

$\therefore$ The time period of the oscillation is 2 seconds .