Answer
given,
wavelength = λ = 18.7 cm
= 0.187 m
amplitude , A = 2.34 cm
v = 0.38 m/s
A) angular frequency = ?
angular frequency ,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) the wave number ,
C)
as the wave is propagating in -x direction, the sign is positive between x and t
y ( x ,t) = A sin(k x - ω t)
y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)
Answer:
(A) The spring constant is <u>98 N/m.</u>
(B) The period of oscillation is <u>0.635 s.</u>
Explanation:
Given:
Mass of the body is,
Extension length of the spring is,
Now, let 'k' be the spring constant.
The force acting on the body is due to gravity only and is equal to its weight.
So, weight of the body is given as:
Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,
Hence, the spring constant is 98 N/m.
(B)
Period of oscillation of a body of spring-mass system in SHM is given as:
Plug in the given values and solve for period 'T'. This gives,
Therefore, the period of oscillation is 0.635 s.
Explanation:
a)
Sum of moments = 0 (Equilibrium)
T . cos (Q)*L = m*g*L/2
b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!
c)