Test Index

Simple Harmonic Motion

Section: Physics

© examsnet.com

Question : 3 of 34

Marks: +1, -0

A spring is stretched by 0.2 m when a mass of 0.5 kg is suspended to it. The time period of the spring when 0.5 kg mass is replaced with a mass of 0.25 kg is suspended to it is

(Acceleration due to gravity

=10\ \mathrm{m}\ \mathrm{s}^{-2}

[AP EAPCET 23 May 2025 S2]

$\;0.628\ \mathrm{s}$
$\;6.28\ \mathrm{s}$
$\;62.8\ \mathrm{s}$
$\;0.0628\ \mathrm{s}$

Solution:

Step 1: Find Spring Constant

(k)

The spring stretches by 0.2 m when a 0.5 kg mass hangs from it.

This means the force from the weight

(m_1 g)

equals the force from the spring

(k x)

:

k x=m_1 g

So,

k=\;\frac{m_1 g}{x}=\;\frac{0.5 \times 10}{0.2}=25\ \mathrm{N/m}

Step 2: Use the New Mass

The new hanging mass is

m_2=0.25\ \mathrm{kg}

.

Step 3: Find the New Time Period (

T

)

The time period of a mass-spring system is:

T=2\pi \sqrt{\frac{m_2}{k}}

Substitute the values:

T=2\pi \sqrt{\frac{0.25}{25}}

T=2\pi \sqrt{0.01}

T=2\pi \times 0.1=0.2\pi\ \mathrm{s}

Step 4: Calculate the Value Using

\pi \approx 3.14

:

T=0.2 \times 3.14=0.628\ \mathrm{s}

© examsnet.com

Go to Question:

Prev Question

Next Question