Question

A $1kg$ mass attached to a spring of force constant $25N/m$ oscillates on a horizontal frictionless track. At $t=0$ the mass is released from rest at $x=-3cm$, i.e. the spring is compressed by $3cm$. Determine velocity as a function of time.

• A. $\left(15cm/s\right)sin\left(\right(5s^{-1}\left)t\right)$
• B. $\left(15cm/s\right)cos\left(\right(5s^{-1}\left)t\right)$
• C. $\left(25cm/s\right)sin\left(\right(3s^{-1}\left)t\right)$
• D. $\left(25cm/s\right)cos\left(\right(3s^{-1}\left)t\right)$

Answer 1

Answer: (A)

$\left(15cm/s\right)sin\left(\right(5s^{-1}\left)t\right)$

For a spring-mass system, the frequency of oscillations is given by $\omega =\sqrt{\frac{k}{m}}$

Given $k=25\text{N/m}$ and $m=1\text{kg}$

Thus, $\omega =\sqrt{25}=5\text{rad/s}$

We know that, the displacement in SHM is represented by $x=A\sin (\omega t+\varphi )$

Thus, velocity is given by $v=A\omega \cos (\omega t+\varphi )$

Given that, at $t=0$ we have $x=-3\text{cm}$ and $v=0$

Thus, $\cos \varphi =0$ and $A\sin \varphi =-3$

Hence, $A=3\text{cm}$ and $\varphi =-\pi /2$

Hence, velocity is civen by $v\left(t\right)=3\times 5\cos (5t-\pi /2)=15\sin \left(5t\right)$