Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

A Wooden Block of Mass 0.5 Kg and Density 800 Kg M−3 is Fastened to the Free End of a Vertical Spring of Spring Constant 50 N M−1 Fixed at the Bottom.Find the Time-period of Vertical Oscillations - Physics

Short Note

A wooden block of mass 0.5 kg and density 800 kg/m3 is fastened to the free end of a vertical spring of spring constant 50 N/m1 fixed at the bottom. If the entire system is completely immersed in water, find the time-period of vertical oscillations of the block when it is slightly depressed and released.

Advertisement Remove all ads

Solution

Given:
Mass of the wooden block, m = 5 kg
Density of the block, ρ = 800 kg/m3
Spring constant, k = 50 N/m
Density of water, ρw = 1000 kg/m3

 As the system is inside the water, the unbalance force will be the driving force, which is kx for SHM.
Hence, there will be no change in the buoyant force.

\[\therefore \text{ ma = kx}\]
\[ \Rightarrow \text{ a = kx/m }\]
\[ \Rightarrow \text{w}^2\text{ x = kx/m }\]
\[ \Rightarrow \left( \frac{2\text{p}}{\text{T}} \right)^2 = \text{ k/m }\]
\[ \Rightarrow \text{T} = 2\pi\sqrt{\frac{\text{m}}{\text{k}}}\]
\[ = 2\pi \times \sqrt{\left( \frac{0 . 5}{50} \right)} = \frac{\pi}{5} \text{s}\] 

Concept: Introduction to Fluid Machanics
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 13 Fluid Mechanics
Q 24.2 | Page 275
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×