Question

An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial, the initial velocity is v0 and in the second it is 4v0. In the second trial, A : the amplitude is twice as great and the maximum acceleration is half as great. B : both the amplitude and the maximum acceleration are four times as great. C : the amplitude is half as great and the maximum acceleration is twice as great. D : both the amplitude and the maximum acceleration are twice as great. E : the amplitude is four times as great and the maximum acceleration is twice as great.

Answer 1

Explanation:

It is given that, in the first trial, the initial velocity is $v_o$ and in the second it is $4v_o$.

The total energy of the system remains constant. So,

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=\text{constant}$ ....(1)

x is amplitude

It means that the amplitude is directly proportional to velocity. If velcoity increases to four times, then the amplitude also becomes 4 times.

Differentiating equation (1) we get :

$mv\dfrac{dv}{dt}+kx\dfrac{dx}{dt}=0$

Since,

$\dfrac{dv}{dt}=a,\ \text{acceleration}$ and $\dfrac{dx}{dt}=v,\ \text{velocity}$

So,

$mva+kxv=0$

It means that the acceleration is also proportional to the amplitude. So, acceleration also becomes 4 times.

Hence, the correct option is (B) "both the amplitude and the maximum acceleration are four times as great"