Question

A simple harmonic oscillator consist of a block attached to a spring with k = 100 N/m. The block slides on a frictionless horizontal surface, with equilibrium point x = 0. A graph of block's velocity as a function of time is shown. correctly match the two column:- (${\pi }^2=10)$
List - I List - II
(P) The block's mass in Kg (1) -0.20
(Q) The block's displacement at t = 0 in meters (2) -200
(R) The block's acceleration at t = 0.1 s in $m/s^2$ (3) 0.20
(S) The block's maximum kinetic energy in joules (4) 4.0

• A. P-2 Q-1 R-4 S-3
• B. P-1 Q-3 R-2 S-4
• C. P-4 Q-2 R-3 S-1
• D. P-3 Q-4 R-1 S-2

Answer 1

Answer: (B)

$\left(P\right)T=2\pi \sqrt{\frac{m}{K}}0.1=2\times 3.14\sqrt{\frac{m}{100}}m={\left(\frac{0.1}{2\times 3.14}\right)}^2\times 100m=0.025Kg$

$\left(Q\right)$ From above graph we can get,

$v=\pi sin\left(\sqrt{\frac{K}{m}}t\right)v=\pi sin\left(\sqrt{\frac{100}{0.025}}t\right)v=\pi sin\left(20\sqrt{10}t\right)\frac{dx}{dt}=\pi sin\left(20\sqrt{10}t\right){\int }_o^{x}dx={\int }_0^t\pi sin\left(20\sqrt{10}t\right)dtx=\frac{\pi }{20\sqrt{10}}cos\left(20\sqrt{10}t\right)$

at $t=0x=-\frac{\pi }{20\sqrt{10}}=0.05m$

$\left(R\right)a=\frac{dv}{dt}=\pi \sqrt{10}\times 20cos\left(20\sqrt{10}t\right)att=0.1a=198.6\approx 200m/s^2$

$\left(S\right)V_{max}=\pi [whensin\theta =1]{KE}_{max}=\frac{1}{2}\times 0.025\times {\pi }^2=0.125Joule$