Question

A $ 100g $ mass stretches a particular spring by $ 9.8cm $ , when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $ 6.28s $ ?
(A) $ 1000g $
(B) $ {10^5}g $
(C) $ {10^7}g $
(D) $ {10^4}g $

Answer 1

Hint : The given data suggests that the spring of initial length is extended by $ 9.8cm $ when a mass of $ 100g $ is suspended vertically on it . Find the spring constant of the spring using hooke’s law equation and use it in the time period of the oscillation equation to find the mass value when subtended on dame spring.

Formula used:
 $ F = Kx $
 $ T = 2\pi \sqrt {\dfrac{M}{K}} $ .

Complete step by step Solution:
Hooke’s law states that the force required to elongate or compress a spring by a distance is directly proportional to the distance of compression or elongation. Removing the proportionality constant we get a constant term called spring constant. Spring constant of a spring is used to determine the value of force applied on the spring and it is a constant factor for all spring. It varies from the type of spring used.
Now , mathematically Hooke’s law equation is given as:
 $ \Rightarrow F = - Kx $
Where F is the total spring force, K is the spring constant and x is the distance of elongation or compression.
From the given data we know that the spring stretched by $ 9.8cm $ , which we can substitute to find spring constant.
 $ \Rightarrow K = \dfrac{F}{x} $
From Newton’s second law, we know that force is a product of its mass and acceleration. Here, when the body is suspended vertically, the acceleration is due to gravity
 $ \Rightarrow K = \dfrac{{mg}}{x} $
Substituting the values we get
 $ \Rightarrow K = \dfrac{{0.1kg \times 9.8}}{{9.8 \times {{10}^{ - 2}}m}} $
On Simplifying we get,
 $ \Rightarrow K = 10 $
Now, when another block of mass M is subtended vertically, there will be vibrations produced by the spring. The time period of the vibrations are given and asked to find out mass. We can find this using time period of SHM formula,
 $ \Rightarrow T = 2\pi \sqrt {\dfrac{M}{K}} $
 $ \Rightarrow 6.28 = 2\pi \sqrt {\dfrac{M}{{10}}} $
On further simplification we get,
 $ \Rightarrow 1 = \sqrt {\dfrac{M}{{10}}} $
Squaring on both sides we get,
 $ \Rightarrow 1 = \dfrac{M}{{10}} $
 $ \Rightarrow M = 10Kg $
Hence, option (D) is the right answer.

Note:
We use a time period of Simple harmonic motion formula for this application because, when a mass is subtended on a spring, the spring undergoes constant to and fro vibrating motion which resembles a SHM type of motion when a graph is plotted against, distance against time.