Question

. An uncalibrated spring balance is found to have a
period of oscillation of 0.314 s, when a 1 kg weight
is suspended from it? How does the spring
elongate, when a 1 kg weight is suspended from it?
[Take 1 = 3.14
(Ans. 2.45 cm)​

Answer 1

Spring is elongated by 2.45 cm

Explanation:

It is given time period of oscillation T = 0.314 sec

Mass = 1 kg

We know that angular frequency $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.314}=20rad/sec$

Now $\omega =\sqrt{\frac{m}{k}}$

So $20=\sqrt{\frac{1}{k}}$

Squaring both side

$400={\frac{1}{k}}$

k = 0.0025

Spring force is given by $F=kx$

This spring force will balance the weight

So $kx=mg$

$0.0025\times x=1\times 9.8$

$x=0.0245m=2.45cm$

Learn more

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displayed and released, oscillates with a period of 0.6 s. What is the weight of the body?

https://brainly.in/question/7382074

Answer 2

Spring Elongation = L = 2.45 cm

Explanation:

Time period = T = 0.314 s

Mass of weight = m = 1 kg

We know that:

T = 2π √(m/K)  

T^2 / 4π^2 = m/K

K =  4*π^2*m/T^2

K = 4π^2(1) / (0.314)^2

K = 400.41 N/m

Spring elongation when 1 kg weight is suspended with the spring can be found as:

Fw = FL

mg = KL

L = mg/K  

L = (1)(9.81)/400.40

L = 0.0245 m

L = 2.45 cm