A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?

Question

A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed? (A) 10 m / s 2 (B) 5 m / s 2 (C) 4 m / s 2 (D) 9.8 m / s 2

Tardigrade Admin · Accepted Answer

From potential energy curve U max =(1/2) kA 2 ⇒ 10=(1/2) k (2)2 ⇒ k =5 Now T text spring = T text pendulum 2 π √(5/5)=2 π √(4/ g ) ⇒ 1=√(4/ g ) ⇒ g =4 on planet

$10 m / s^{2}$

$5 m / s^{2}$

$4 m / s^{2}$

$9.8 m / s^{2}$

From potential energy curve
$U_{m a x} = \frac{1}{2} k A^{2}$
$\Rightarrow 10 = \frac{1}{2} k (2)^{2}$
$\Rightarrow k = 5$
Now $T_{spring} = T_{pendulum}$
$2 π \frac{5}{5} = 2 π \frac{4}{g}$
$\Rightarrow 1 = \frac{4}{g}$
$\Rightarrow g = 4$ on planet

10m/s2

5m/s2

4m/s2

9.8m/s2

From potential energy curve Umax​=21​kA2 ⇒10=21​k(2)2 ⇒k=5 Now Tspring ​=Tpendulum ​ 2π55​​=2πg4​​ ⇒1=g4​​ ⇒g=4 on planet

$10 m / s^{2}$

$5 m / s^{2}$

$4 m / s^{2}$

$9.8 m / s^{2}$

From potential energy curve
$U_{m a x} = \frac{1}{2} k A^{2}$
$\Rightarrow 10 = \frac{1}{2} k (2)^{2}$
$\Rightarrow k = 5$
Now $T_{spring} = T_{pendulum}$
$2 π \frac{5}{5} = 2 π \frac{4}{g}$
$\Rightarrow 1 = \frac{4}{g}$
$\Rightarrow g = 4$ on planet