The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
Answer: C. Steel
Explanation: When a sound wave travels through a solid body consisting
of an elastic material, the velocity of the wave is relatively
high. For instance, the velocity of a sound wave traveling
through steel (which is almost perfectly elastic) is about
5,060 meters per second. On the other hand, the velocity
of a sound wave traveling through an inelastic solid is
relatively low. So, for example, the velocity of a sound wave
traveling through lead (which is inelastic) is approximately
1,402 meters per second.
The answer is the first one
Answer:
k = 4422.35 KN/m
Explanation:
Given that
Frequency ,f= 29 Hz
m = 7.5 g
Natural frequency ω
ω = 2 π f
We also know that for spring mass system
ω ² m =k
k=Spring constant
So we can say that
( 2 π f)² = m k
By putting the values
(2 x π x 29)² = 7.5 x 10⁻³ k
33167.69 = 7.5 x 10⁻³ k
k=4422.35 x 10³ N/m
k = 4422.35 KN/m
Therefore spring constant will be 4422.35 KN/m
