We know that in a simple harmonic motion of a spring the frequency depends only on spring constant(k) and mass stuck to spring (m) and is given by formula

$f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}}$

From the formula we can see for a fixed spring and mass frequency doesn't change. But the velocity of mass at any time is given by formula

$v(t)={\frac {\mathrm {d} x}{\mathrm {d} t}}=-A\omega \sin(\omega t-\varphi )$

where ${\displaystyle \omega ={\sqrt {\frac {k}{m}}},\qquad A={\sqrt {{c_{1}}^{2}+{c_{2}}^{2}}},\qquad \tan \varphi ={\frac {c_{2}}{c_{1}}},}$

where $c_1,c_2$ are constants which depend on initial conditons.

Now,if we talk only about the magnitude of velocity(speed) then we can increase the maximum speed by increasing values of $c_1,c_2$ by some means and proportionately the speed at points between the two extreme points will also increase(not sure).

So,my question is if we are increasing the speed at each point where is neutralization taking place so that the frequency(constant) doesn't increase,because if we increase the speed at each point frequency should also increase?

I hope you got my question.


If you increase the velocity you also increase the amplitude so although the object might be travelling faster at times it will have to travel further to complete an oscillation.

It is a characteristic of simple harmonic motion that the linkage between amplitude $A$ and maximum velocity $v_{\rm max}$ is such that the period of the motion $T$ does not change $v_{\rm max}=A\omega=A\dfrac{2\pi}{T}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.