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:- ( $π^{2} = 10)$
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

Question

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Answer 1

$(P) T = 2 π \frac{m}{K} 0.1 = 2 \times 3.14 \frac{m}{1 0 0} m = (\frac{0 . 1}{2 \times 3 . 1 4})^{2} \times 100 m = 0.025 K g$

$(Q)$ From above graph we can get,
$v = π s i n (\frac{K}{m} t) v = π s i n (\frac{1 0 0}{0 . 0 2 5} t) v = π s i n (2010 t) \frac{d x}{d t} = π s i n (2010 t) \int_{o}^{x} d x = \int_{0}^{t} π s i n (2010 t) d t x = \frac{π}{2 0 1 0} c o s (2010 t)$
at $t = 0 x = - \frac{π}{2 0 1 0} = 0.05 m$

$(R) a = \frac{d v}{d t} = π 10 \times 20 c o s (2010 t) a t t = 0.1 a = 198.6 \approx 200 m / s^{2}$

$(S) V_{m a x} = π [w h e n s i n θ = 1] K E_{m a x} = \frac{1}{2} \times 0.025 \times π^{2} = 0.125 J o u l e$

Answer 2

(P) T = 2 π \frac{m}{K} 0.1 = 2 \times 3.14 \frac{m}{1 0 0} m = (\frac{0 . 1}{2 \times 3 . 1 4})^{2} \times 100 m = 0.025 K g

(Q)

From above graph we can get,

v = π s i n (\frac{K}{m} t) v = π s i n (\frac{1 0 0}{0 . 0 2 5} t) v = π s i n (2010 t) \frac{d x}{d t} = π s i n (2010 t) \int_{o}^{x} d x = \int_{0}^{t} π s i n (2010 t) d t x = \frac{π}{2 0 1 0} c o s (2010 t)

at

t = 0 x = - \frac{π}{2 0 1 0} = 0.05 m

(R) a = \frac{d v}{d t} = π 10 \times 20 c o s (2010 t) a t t = 0.1 a = 198.6 \approx 200 m / s^{2}

(S) V_{m a x} = π [w h e n s i n θ = 1] K E_{m a x} = \frac{1}{2} \times 0.025 \times π^{2} = 0.125 J o u l e