The variations of potential energy (U) with position x for three simple harmonic oscillators A, B and C are shown in figure. The oscillators have same mass. The time period of oscillation is greatest for

Question

The variations of potential energy (U) with position x for three simple harmonic oscillators A, B and C are shown in figure. The oscillators have same mass. The time period of oscillation is greatest for (A) A (B) B (C) C (D) Same for all

Tardigrade Admin · Accepted Answer

U=(1/2) k x2 x2=(2 U/k) or x propto (1/k) (Since U is constant) Also T=2 π √(m/k) or T propto (1/√k) Therefore x propto T Hence the oscillation with maximum x will have the maximum time period.

Q. The variations of potential energy $(U)$ with position $x$ for three simple harmonic oscillators $A, B$ and $C$ are shown in figure. The oscillators have same mass. The time period of oscillation is greatest for

A

B

C

Same for all

$U = \frac{1}{2} k x^{2}$
$x^{2} = \frac{2 U}{k}$
or $x \propto \frac{1}{k}$ (Since $U$ is constant)
Also $T = 2 π \frac{m}{k}$
or $T \propto \frac{1}{k}$
Therefore $x \propto T$
Hence the oscillation with maximum $x$ will have the maximum time period.

Q. The variations of potential energy (U) with position x for three simple harmonic oscillators A,B and C are shown in figure. The oscillators have same mass. The time period of oscillation is greatest for

A

B

C

Same for all

U=21​kx2 x2=k2U​ or x∝k1​ (Since U is constant) Also T=2πkm​​ or T∝k​1​ Therefore x∝T Hence the oscillation with maximum x will have the maximum time period.

Q. The variations of potential energy $(U)$ with position $x$ for three simple harmonic oscillators $A, B$ and $C$ are shown in figure. The oscillators have same mass. The time period of oscillation is greatest for

$U = \frac{1}{2} k x^{2}$
$x^{2} = \frac{2 U}{k}$
or $x \propto \frac{1}{k}$ (Since $U$ is constant)
Also $T = 2 π \frac{m}{k}$
or $T \propto \frac{1}{k}$
Therefore $x \propto T$
Hence the oscillation with maximum $x$ will have the maximum time period.