Question

The time period of oscillations of mass 0.1 kg suspended from a Hooke’s
spring is 1s. Calculate the time period of oscillation of mass 0.9 kg when
suspended from the same spring

Answer 1

Answer:

w (angular frequency) for SHM = (k / m)^1/2

w2 / w1 = (m1 / m2)^1/2 = (1 / 9)^1/2 = 1 / 3

Since T = 1/f = 1 / ( 2 * pi * w)

the time period for the larger mass is 3 times that of the smaller mass

or 3 sec

Answer 2

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.

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