Question

A block of mass $m=4$ kg undergoes simple harmonic motion with amplitude $A=6$ cm on the frictionless surface. Block is attached to a spring of force constant $k=400N/m$. If the block is at $x=6$ cm at time $t=0$ and equilibrium position is at $x=0$ then the blocks position as a function of time (with $x$ in centimetres and $t$ in seconds)?

• A. $x=6sin(10t+\frac{1}{2}\pi )$
• B. $x=6sin\left(10\pi t\right)$
• C. $x=6sin(10\pi t-\frac{1}{2}\pi )$
• D. $x=6sin(10t-\frac{1}{4}\pi )$

Answer 1

The correct option is A $x=6sin(10t+\frac{1}{2}\pi )$

The position of block executing SHM is $x=A\sin (\frac{2\pi t}{T}+\varphi )$

where $\varphi$ is the initial phase

given at $t=0$ , $x=6cm$ and $A=6cm$

therefore by above equation $6=6\sin (0+\varphi )$

or $1=\sin \varphi$

or $\sin \frac{\pi }{2}=\sin \varphi$

or $\varphi =\frac{\pi }{2}$

now time period of system $T=2\pi \sqrt{\frac{m}{k}}$

$T=2\pi \sqrt{\frac{4}{400}}$

$T=\pi /5$

so position of block $x=6\sin (10t+\frac{\pi }{2})$

as the initial phase is in positive x-direction therefore $+$ sign is taken there.