m is the mass of the body;
g is the acceleration due to gravity;
h is the height of the body above the reference plane;
The reference plane plays a vital role in defining the potential energy of a body. Most often reference
plane is taken as the ground and hence plane B is the reference plane here. We will discuss potential
energy of the body for all the planes individually.
For plane A, the height of the body is zero as it is kept on the same plane A. Hence, putting the value
of h in equation 1 we get
PE = 0
Therefore, potential energy of the body is zero with respect to the plane A. Thus, A is an incorrect
option.
Both the planes A and C are at same height above the reference plane. Therefore, potential energy of
the body with respect to the plane C is same as that in respect of A. Hence, B is incorrect as well.
Since the plane A is at some height from the reference plane, h will have a non-zero value. Therefore,
potential energy of the body with respect to the ground plane will have some non-zero value. Thus, C
is the correct answer option. In the view of the above, D and E are also incorrect options.
Question 7
Displacement of a particle in SHM is given as
x(t) = 23 sin(314t) cm
What is the amplitude and frequency of oscillation of the particle?
A. 2.6 cm, 40 Hz
B. 2.3 cm, 40 Hz
C. 2.3 cm, 50 Hz
D. 2.6 cm, 50 Hz
E. 3.14 cm, 314 Hz
Correct Answer: C
Explanation:
General expression for displacement of a body in SHM is given as
X(t)
= A sin(ωt) ----------- (1)
Where
X(t) is the displacement of the particle;
A is the amplitude of the oscillation;
ω is the angular frequency;
Displacement of a particle as given in the question is
X(t) = 23 sin(314t) cm ------------- (2)
Comparing equation 1 with 2 we get
A = 2.3 cm and ω = 314 rad/s
Angular frequency is related to frequency as ω = 2Πf ----------- (3)
Where
f is the frequency of the oscillation;
Putting the value of in equation 3 we get
314 = 2Πf
Or, 314/(2*Π) = f
Or, 314/(2*3.14) = f
(Π = 3.14)
Or, f = 50 Hz
