Topic: Theory of Machines and Vibrations
A block whose mass is 650 gm is fastened to a spring of spring constant K equals 65 $N/m$ whose other end is fixed. The block is pulled a distance $x=11cm$ from its equilibrium position at $x=0$ on a smooth surface, and released from rest at $t=0$ The maximum speed 'S' of the oscillating block is
- A. $11~cm/sec$
- B. $11~m/sec$
- C. $11~mm/sec$
- D. $1.1~m/sec$
Answer: D. $1.1~m/sec$
Explanation: Given that, mass $(m)=650g=0.65~kg$, $k=65~N/m$, $x=0.11m$. Maximum speed of the spring will be at $x=0$ when the total spring energy is converted into kinetic energy of mass. $\frac{1}{2}kx^{2}=\frac{1}{2}mV^{2} \implies \frac{1}{2}\times65\times(0.11)^{2}=\frac{1}{2}\times(0.650)V^{2} \implies V=1.1~m/sec.$