Question

If amplitude of a spring block system executing SHM is halved, then

• A. Time period halves
• B. Maximum acceleration halves
• C. Maximum speed doubles
• D. Maximum kinetic energy halves

Answer 1

The correct option is B Maximum acceleration halves

Let equation of SHM $x=Asin\left(\omega t\right)$

Velocity $v=A\omega cos\left(\omega t\right)$ $\Rightarrow v_{max}=A\omega$

Maximum Kinetic energy $=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2$

Acceleration $a=-A{\omega }^{2}sin\left(\omega t\right)$ $\Rightarrow a_{max}=A{\omega }^2$

Time Period $=\frac{2\pi }{\omega }$

Now amplitude is halved.

(a) Time Period is independent of Amplitude. So time period is remains same.

(b) Maximum acceleration $=A{\omega }^2$ So new acceleration is $a^{\prime }=\frac{1}{2}A{\omega }^2$

(c) Maximum speed $v=A\omega$ so new speed is $v^{\prime }=\frac{1}{2}A\omega$

(d) Maximum Kinetic energy $=\frac{1}{2}m{A}^2{\omega }^2$ So new kinetic energy is $K{E}^{\prime }=\frac{1}{8}m{A}^2{\omega }^2$