Question

A block of mass $2kg$ is attached to the spring of spring constant $50N/m$ . The block is pulled to a distance of $5cm$ from its equilibrium position $\left(\text{at }x=0\right)$ on a horizontal frictionless surface and released at $t=0$ from rest. The expression for its displacement at anytime $t$ is

• A. $5sin\left(5t+\pi /2\right)$
• B. $sin\left(5t+\pi /2\right)$
• C. $5sin\left(5t+3\pi /2\right)$
• D. $5sin\left(t+\pi /2\right)$

Answer 1

Answer: (A)

The spring -block system will perform SHM about the mean position with an amplitude $5cm$ .

Given, spring constant $k=50N/m$
m = mass attached = 2 kg
$\therefore \text{Angular frequency}\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5\text{rad/s}$
Assuming the displacement function
y(t) = A sin (ωt + ϕ)
where, ϕ = initial phase
But given at t = 0, y(t) = +A
∴ y(0) = + A = A sin (ω × 0 + ϕ)
or $\text{sin}\varphi =1\Rightarrow \varphi =\frac{\pi }{2}$
& the desired equation is
$y\left(t\right)=A\text{sin}\left(\omega t+\frac{\pi }{2}\right)$
= A cos ωt
Putting A = 5 cm, ω = 5 rad/s
We get, y(t) = 5 sin (5t + π/2)
where, t is in second and y is in centimetre.